The first branch points at the body
|| and && compile to the same two compares over the same two values. What separates them is where the first branch points.
0 cmp r0, #0
2 blt 8 ~>
4 cmp r1, #0
6 bge 14 ~>
8 ~>mov r0, #1
10 neg r0, r0
12 b 16 ~>
14 ~>add r0, r1
16 ~>bx lr
The branch at 2 jumps into the body at 8, and it does so with the condition exactly as the source wrote it — one test passing is enough, so there is nothing left to check. The branch at 6 is the last one, and it is the inverted one: with no further tests to fall back on, failing it means the whole condition failed, so it jumps past the body to 14.
Now the same function with the operator changed and nothing else touched:
0 cmp r0, #0
2 bge 14 ~>
4 cmp r1, #0
6 bge 14 ~>
8 mov r0, #1
10 neg r0, r0
12 b 16 ~>
14 ~>add r0, r1
16 ~>bx lr
Both branches inverted, both pointing at 14. Two compares, two branches, the same body, the same eighteen bytes — and the operator is legible purely from whether the branch targets agree.
The body in both is worth a glance on its way past: mov r0, #1 followed by neg r0, r0. Minus one has no single-instruction Thumb encoding, so the compiler builds it in two whenever it has to materialise it fresh on a return path.
Your target uses the or-shape with a different relation in both tests. Read the first branch's destination to confirm which operator you are looking at, then invert the last one to recover the second test.
Your task
Write func_0815471c to reproduce the target assembly.