This problem was just an exercise in writing assembly code with only 2 temporary registers (technically the J register was a temporary until the program was done). The logic of my program was just if any of the next 3 spots (A, B, or C) was a hole, and the tile 4 spots away (D where you land if you jump immediately) wasn't a hole, then jump.
NOT A T
OR T J
// J = !A
NOT B T
OR T J
// J = !A || !B
NOT C T
OR T J
// J = !A || !B || !C
AND D J
// J = (!A || !B || !C) && D
WALK
I could have cut out the second instruction by just making the first NOT write to J.
Part 2 just used a harder sequence of holes that required you too look ahead further, up to 9 spots (which new read-only registers E through I). The logic of my program was to start with my solution for part 1 (if a hole in A B or C and ground at D), and add in the condition that either there must be ground on the tile after I land (E) or if I needed to jump again immediately, the tile I landed on after that (H) was ground. This avoids the case where you jump sooner than needed (because there is a hole at C) only to find that your only choice is to jump into a 3 wide pit. There are likely cases that could be possible to solve with only 15 instructions that this doesn't cover, but it worked for me.
NOT A T
OR T J
NOT B T
OR T J
NOT C T
OR T J
AND D J
// J = (!A || !B || !C) && D
NOT J T
AND J T
// the two above instructions set T to false so I can OR into it
OR E T
OR H T
AND T J
// J = (!A || !B || !C) && D && (E || H)
RUN