05AB1E, 77 74 bytes

0V‘‚µ‘©IÇv¼®¾4%3÷׫Yy₁+b¦RCDV-₁%¾‘DO,1—¨#ÿ<-#ÿ‘«}¾‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾

Port of @Lynn's Python answer, so make sure to upvote him/her as well!

Outputs with spaces at 1 SUB and DO READ OUT,1 DO GIVE UP.

Try it online or verify a few more test cases.

Try the output in INTERCAL.

0V                 # Set variable `Y` to 0 (it's 2 by default)
‘‚µ‘               # Push dictionary string "PLEASE"
    ©              # Store it in variable `®` (without popping)
IÇ                 # Push the input, and convert it to a list of codepoint integers
  v                # Loop over each integer `y`:
   ¼               #  Increment the counter_variable (0 by default)
    ¾              #  Push the counter_variable
     4%            #  Modulo-4
       3÷          #  Integer-divided by 3
   ®     ×         #  Repeat "PLEASE" from variable `®` that many times
          «        #  Append it to the string
   Y               #  Push the current `Y`
    y              #  Push the current codepoint
     ₁+            #  Increase it by 256
       b           #  Convert it to a binary string
        ¦          #  Remove the leading 1 (the `₁+` and `¦` are to pad leading 0s)
         R         #  Reverse it
          C        #  Convert it from binary to a base-10 integer
           DV      #  Store a copy as new value for variable `Y`
           -       #  Subtract it from the `Y` we've pushed earlier
            ₁%     #  Modulo-256
   ¾               #  Push the counter_variable
    ‘DO,1—¨#ÿ<-#ÿ‘ #  Push dictionary string "DO,1 SUB#ÿ<-#ÿ", where the `ÿ` are 
                   #  automatically filled with the counter_variable and `Y`%256
      «            #  Append it to the string as well
}¾                 # After the loop: push the counter_variable
  ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾
                   # Push dictionary string "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP", where
                   # the `ÿ` are automatically filled with the counter_variable and
                   # string
                   # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ‘‚µ‘ is "PLEASE"; ‘DO,1—¨#ÿ<-#ÿ‘ is "DO,1 SUB#ÿ<-#ÿ"; and ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾ is "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP".

05AB1E, 77 74 54 bytes

Ç₁+₁šb퀨C¥(₁%€‘‚µ,1<-#1€·,1—¨#1<-#ÿ€·‚Ø€Ä,1‘J‘ÿ€·†¿€¾

-20 bytes by porting @UnrelatedString's Jelly answer, so make sure to upvote him!!

Outputs with spaces at #1 DO,1 SUB#, DO READ OUT, and DO GIVE UP.

Try it online or verify a few more test cases.

Try the output in INTERCAL.

Ç               # Convert the (implicit) input-string to a list of codepoint integers
 ₁+             # Add 256 to each value
   ₁š           # Prepend 256 at the front of the list
     b          # Convert each value to a binary string
      í         # Reverse each
       ۬       # Remove the last digit of each
         C      # Convert each from binary back to a base-10 integer
          ¥     # Take the forward differences (deltas) of this list
           (    # Negate each difference
            ₁%  # Take modulo-256 on each
€               # Map over each integer:
 ‘‚µ,1<-#1€·,1—¨#1<-#ÿ€·‚Ø€Ä,1‘
                #  Push dictionary string "PLEASE,1<-#1 DO,1 SUB#1<-#ÿ DO READ OUT",
                #  where the `ÿ` is automatically filled with the integer
   J            # Join this list of strings together
    ‘ÿ€·†¿€¾    # Push string "ÿ DO GIVE UP", where the `y` is filled with the string
                # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ‘‚µ,1<-#1€·,1—¨#1<-#ÿ€·‚Ø€Ä,1‘ is "PLEASE,1<-#1 DO,1 SUB#1<-#ÿ DO READ OUT" and ‘ÿ€·†¿€¾ is "ÿ DO GIVE UP".

05AB1E, 77 bytes

0‘‚µ‘©IÇv¼sDy₁+b¦RC-DV-s®¾4%3÷׫Y₁%¾‘DO,1—¨#ÿ<-#ÿ‘«}¾‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾

Try it online or verify a few more test cases.

0                # Push 0 (head-integer)
‘‚µ‘             # Push dictionary string "PLEASE"
    ©            # Store it in variable `®` (without popping)
IÇ               # Push the input, and convert it to a list of codepoint integers
  v              # Loop over each integer `y`:
   ¼             #  Increment the counter_variable (0 by default)
   s             #  Swap so the head-integer is at the top of the stack
    D            #  Duplicate it
     y           #  Push the current codepoint
      ₁+         #  Increase it by 256
        b        #  Convert it to a binary string
         ¦       #  Remove the leading 1 (the `₁+` and `¦` are to pad leading 0s)
          R      #  Reverse it
           C     #  Convert it from binary to a base-10 integer
            -    #  Subtract it from the duplicated head-integer
             DV  #  Store a copy in variable `Y`
               - #  Subtract it from the head-integer
   s             #  Swap so the string is at the top again
     ¾           #  Push the counter_variable
      4%         #  Modulo-4
        3÷       #  Integer-divided by 3
    ®     ×      #  Repeat "PLEASE" from variable `®` that many times
           «     #  Append it to the string
    Y            #  Push integer `Y` again
     ₁%          #  Modulo-256
       ¾         #  Push the counter_variable
        ‘DO,1—¨#ÿ<-#ÿ‘
                 #  Push dictionary string "DO,1 SUB#ÿ<-#ÿ", where the `ÿ` are 
                 #  automatically filled with the counter_variable and `Y`%256
          «      #  Append it to the string as well
}¾               # After the loop: push the counter_variable
  ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾
                 # Push dictionary string "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP", where
                 # the `ÿ` are automatically filled with the counter_variable and
                 # string
                 # (after which the result is output implicitly)

05AB1E, 77 74 bytes

0V‘‚µ‘©IÇv¼®¾4%3÷׫Yy₁+b¦RCDV-₁%¾‘DO,1—¨#ÿ<-#ÿ‘«}¾‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾

Try it online or verify a few more test cases.

0V                 # Set variable `Y` to 0 (it's 2 by default)
‘‚µ‘               # Push dictionary string "PLEASE"
    ©              # Store it in variable `®` (without popping)
IÇ                 # Push the input, and convert it to a list of codepoint integers
  v                # Loop over each integer `y`:
   ¼               #  Increment the counter_variable (0 by default)
    ¾              #  Push the counter_variable
     4%            #  Modulo-4
       3÷          #  Integer-divided by 3
   ®     ×         #  Repeat "PLEASE" from variable `®` that many times
          «        #  Append it to the string
   Y               #  Push the current `Y`
    y              #  Push the current codepoint
     ₁+            #  Increase it by 256
       b           #  Convert it to a binary string
        ¦          #  Remove the leading 1 (the `₁+` and `¦` are to pad leading 0s)
         R         #  Reverse it
          C        #  Convert it from binary to a base-10 integer
           DV      #  Store a copy as new value for variable `Y`
           -       #  Subtract it from the `Y` we've pushed earlier
            ₁%     #  Modulo-256
   ¾               #  Push the counter_variable
    ‘DO,1—¨#ÿ<-#ÿ‘ #  Push dictionary string "DO,1 SUB#ÿ<-#ÿ", where the `ÿ` are 
                   #  automatically filled with the counter_variable and `Y`%256
      «            #  Append it to the string as well
}¾                 # After the loop: push the counter_variable
  ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾
                   # Push dictionary string "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP", where
                   # the `ÿ` are automatically filled with the counter_variable and
                   # string
                   # (after which the result is output implicitly)

Port of @Lynn's Python answer, so make sure to upvote him as well!

0                # Push 0 (head-integer)
‘‚µ‘             # Push dictionary string "PLEASE"
    ©            # Store it in variable `®` (without popping)
IÇ               # Push the input, and convert it to a list of codepoint integers
  v              # Loop over each integer `y`:
   ¼             #  Increment the counter_variable (0 by default)
   s             #  Swap so the head-integer is at the top of the stack
    D            #  Duplicate it
     y           #  Push the current codepoint
      ₁+         #  Increase it by 256
        b        #  Convert it to a binary string
         ¦       #  Remove the leading 1 (the `₁+` and `¦` are to pad leading 0s)
          R      #  Reverse it
           C     #  Convert it from binary to a base-10 integer
            -    #  Subtract it from the duplicated head-integer
             DV  #  Store a copy in variable `Y`
               - #  Subtract it from the head-integer
   s             #  Swap so the string is at the top again
     ¾           #  Push the counter_variable
      4%         #  Modulo-4
        3÷       #  Integer-divided by 3
    ®     ×      #  Repeat "PLEASE" from variable `®` that many times
           «     #  Append it to the string
    Y            #  Push integer `Y` again
     ₁%          #  Modulo-256
       ¾         #  Push the counter_variable
        ‘DO,1—¨#ÿ<-#ÿ‘
                 #  Push dictionary string "DO,1 SUB#ÿ<-#ÿ", where the `ÿ` are 
                 #  automatically filled with the counter_variable and `Y`%256
          «      #  Append it to the string as well
}¾               # After the loop: push the counter_variable
  ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾
                 # Push dictionary string "DO,1<-$ÿÿ DO READ OUT,1 DO GIVE UP", where
                 # the `ÿ` are automatically filled with the counter_variable and
                 # string
                 # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ‘‚µ‘ is "PLEASE"; ‘DO,1—¨#ÿ<-#ÿ‘ is "DO,1 SUB#ÿ<-#ÿ"; and ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾ is "DO,1<-$ÿÿ DO READ OUT,1 DO GIVE UP".

Port of @Lynn's Python answer, so make sure to upvote him/her as well!

0                # Push 0 (head-integer)
‘‚µ‘             # Push dictionary string "PLEASE"
    ©            # Store it in variable `®` (without popping)
IÇ               # Push the input, and convert it to a list of codepoint integers
  v              # Loop over each integer `y`:
   ¼             #  Increment the counter_variable (0 by default)
   s             #  Swap so the head-integer is at the top of the stack
    D            #  Duplicate it
     y           #  Push the current codepoint
      ₁+         #  Increase it by 256
        b        #  Convert it to a binary string
         ¦       #  Remove the leading 1 (the `₁+` and `¦` are to pad leading 0s)
          R      #  Reverse it
           C     #  Convert it from binary to a base-10 integer
            -    #  Subtract it from the duplicated head-integer
             DV  #  Store a copy in variable `Y`
               - #  Subtract it from the head-integer
   s             #  Swap so the string is at the top again
     ¾           #  Push the counter_variable
      4%         #  Modulo-4
        3÷       #  Integer-divided by 3
    ®     ×      #  Repeat "PLEASE" from variable `®` that many times
           «     #  Append it to the string
    Y            #  Push integer `Y` again
     ₁%          #  Modulo-256
       ¾         #  Push the counter_variable
        ‘DO,1—¨#ÿ<-#ÿ‘
                 #  Push dictionary string "DO,1 SUB#ÿ<-#ÿ", where the `ÿ` are 
                 #  automatically filled with the counter_variable and `Y`%256
          «      #  Append it to the string as well
}¾               # After the loop: push the counter_variable
  ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾
                 # Push dictionary string "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP", where
                 # the `ÿ` are automatically filled with the counter_variable and
                 # string
                 # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ‘‚µ‘ is "PLEASE"; ‘DO,1—¨#ÿ<-#ÿ‘ is "DO,1 SUB#ÿ<-#ÿ"; and ‘DO,1<-#ÿÿ€·‚Ø€Ä,1€·†¿€¾ is "DO,1<-#ÿÿ DO READ OUT,1 DO GIVE UP".