Gol><>, 40 bytes

`e2RFL3%zR"zziF"L5%zR"zzuB"lQlRoaoC|LN|;

Updated for 0.4.0! I'm still tinkering with loops and trying to figure out how to do things best, but this is looking good so far.

Try it online.

Explanation

`e            Push 'e', or 101
2RF ... |     Execute F for loop twice - the first time activates the loop, and the
              second time updates it. This effectively makes the loop start from 1
L3%z          Push 1 if loop counter % 3 is 0, else 1
R"zziF"       Push "Fizz" (top of stack) number of times
L5%z          Push 1 if loop counter % 5 is 0, else 1
R"zzuB"       Push "Buzz" (top of stack) number of times
lQ ... |      If the stack is not empty...
  lRo         Output stack
  ao          Output newline
  C           Continue for loop
LN            Otherwise, print loop counter with newline

;             Terminate program

As we can see, there's a lot of abuse of R, which pops the top of the stack and executes the next instruction that many times.

Gol><>, 40 bytes

`e2RFL5%zR"zzuB"L3%zR"zziF"lQlRoaoC|LN|;

Updated for 0.4.0! I'm still tinkering with loops and trying to figure out how to do things best, but this is looking good so far.

Try it online.

Explanation

`e            Push 'e', or 101
2RF ... |     Execute F for loop twice - the first time activates the loop, and the
              second time updates it. This effectively makes the loop start from 1
L5%z          Push 1 if loop counter % 5 is 0, else 1
R"zzuB"       Push "Buzz" (top of stack) number of times
L3%z          Push 1 if loop counter % 3 is 0, else 1
R"zziF"       Push "Fizz" (top of stack) number of times
lQ ... |      If the stack is not empty...
  lRo         Output stack
  ao          Output newline
  C           Continue for loop
LN            Otherwise, print loop counter with newline

;             Terminate program

As we can see, there's a lot of abuse of R, which pops the top of the stack and executes the next instruction that many times.

Gol><>, 45 bytes

0PV0`d)?;05%z6Q"zzuB"03%z6Q"zziF"lz3Q0NtlRoao

Try it online.

This answer is valid as of 0.3.11. New features might change things/make things golfier in the future, but I think this is pretty good with current features.

Explanation

0              First iteration: push 0. Every iteration after: push n-1.
P              Increment
V0             Save n in variable 0 (almost any char can be a variable name)
`d)?;          Halt if > 100
05%z6Q"zzuB"   Push "Buzz" if n%5 == 0
03%z6Q"zziF"   Push "Fizz" if n%3 == 0
lz3Q0Nt        If the stack is empty, print n and continue loop with t
lRoao          Otherwise, print the stack followed by a newline and continue loop

Gol><>, 40 bytes

`e2RFL3%zR"zziF"L5%zR"zzuB"lQlRoaoC|LN|;

Updated for 0.4.0! I'm still tinkering with loops and trying to figure out how to do things best, but this is looking good so far.

Try it online.

Explanation

`e            Push 'e', or 101
2RF ... |     Execute F for loop twice - the first time activates the loop, and the
              second time updates it. This effectively makes the loop start from 1
L3%z          Push 1 if loop counter % 3 is 0, else 1
R"zziF"       Push "Fizz" (top of stack) number of times
L5%z          Push 1 if loop counter % 5 is 0, else 1
R"zzuB"       Push "Buzz" (top of stack) number of times
lQ ... |      If the stack is not empty...
  lRo         Output stack
  ao          Output newline
  C           Continue for loop
LN            Otherwise, print loop counter with newline

;             Terminate program

As we can see, there's a lot of abuse of R, which pops the top of the stack and executes the next instruction that many times.

Gol><>, 44 bytes

0PV0`d)?;05%z6Q"zzuB"03%z6Q"zziF"lz3Q0NtlRoao

Try it online.

This answer is valid as of 0.3.11. New features might change things/make things golfier in the future, but I think this is pretty good with current features.

Explanation

0              First iteration: push 0. Every iteration after: push n-1.
P              Increment
V0             Save n in variable 0 (almost any char can be a variable name)
`d)?;          Halt if > 100
05%z6Q"zzuB"   Push "Buzz" if n%5 == 0
03%z6Q"zziF"   Push "Fizz" if n%3 == 0
lz3Q0Nt        If the stack is empty, print n and continue loop with t
lRoao          Otherwise, print the stack followed by a newline and continue loop

Gol><>, 45 bytes

0PV0`d)?;05%z6Q"zzuB"03%z6Q"zziF"lz3Q0NtlRoao

Try it online.

This answer is valid as of 0.3.11. New features might change things/make things golfier in the future, but I think this is pretty good with current features.

Explanation

0              First iteration: push 0. Every iteration after: push n-1.
P              Increment
V0             Save n in variable 0 (almost any char can be a variable name)
`d)?;          Halt if > 100
05%z6Q"zzuB"   Push "Buzz" if n%5 == 0
03%z6Q"zziF"   Push "Fizz" if n%3 == 0
lz3Q0Nt        If the stack is empty, print n and continue loop with t
lRoao          Otherwise, print the stack followed by a newline and continue loop