The second sha256 hash is only hashing 32 bytes (the result of the first sha256 hash) so there are only 64 rounds for that.

I just mentioned the 128 rounds to calculate the % reduction in overall calculation, 3 out of all 128 rounds (ignoring the other calculations that need to be done).

Note that the difficulty IS NOT determined by "number of 0's" but rather that the target must be less than a certain value. This happens to result in the hash having a lot of 0's in front of it, but that is not actually what is being checked.

Left this out for simplicity

Also, the hash that is actually produced is the byteswapped version of what we actually see presented to us. I'm not sure if that effects this.

This was factored in.