Haskell, 52 bytes

f k=[(' '<$['B'..s])++(['A'..s]>>s:" ")|s<-['A'..k]]

Returns a list of lines.

Try it online!

f k=                  -- main function is f, input parameter k
  [   |s<-['A'..k]]   -- foreach s from ['A'..k]
   (' '<$['B'..s])    -- replace each element in ['B'..s] with a space
     ++               -- and append
    ['A'..s]>>s:" "   -- (length of ['A'..s]) copies of the two character list
                      -- [s,space]

Haskell, 52 44 bytes

f k=[[" ",s:" "]>>=(['A'..s]>>)|s<-['A'..k]]

Returns a list of lines.

Try it online!

f k=                  -- main function is f, input parameter k
  [   |s<-['A'..k]]   -- for each s from ['A'..k]
     >>=              -- map (and collect the results in a single string) the function: 
         (['A'..s]>>) --  replace each element in ['A'..s] with
    [  ,  ]           --  over the list, containing
     " "              --   a single space to get the indent
        s:" "         --   s followed by space to get the letter sequence

Edit: @jferard: saved three bytes. Thanks!

Haskell, 52 bytes

f k=[(' '<$['B'..s])++(['A'..s]>>s:" ")|s<-['A'..k]]

Try it online!

Haskell, 52 bytes

f k=[(' '<$['B'..s])++(['A'..s]>>s:" ")|s<-['A'..k]]

Returns a list of lines.

Try it online!

f k=                  -- main function is f, input parameter k
  [   |s<-['A'..k]]   -- foreach s from ['A'..k]
   (' '<$['B'..s])    -- replace each element in ['B'..s] with a space
     ++               -- and append
    ['A'..s]>>s:" "   -- (length of ['A'..s]) copies of the two character list
                      -- [s,space]