How to show that FPATH is in NL?
Turning my comment into an answer for the question in the title, the Immerman-Szelepcsényi Theorem says that NL = coNL.  This means it's enough to show that FPATH is in coNL.  For that, observe that the path $a_1,a_2,\ldots, a_n$ is not in $G$ if and only if some edge $a_ia_{i+1}$ is not present.  Since you just need to guess $i$, that's all you need to write down.