How does this proof of the regular dodecahedron's existence fail?

To find the flaw, tile the plane with regular hexagons and apply the “proof” to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.

Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$”; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$” is false for hexagons, and there's no justification why it should be true for pentagons.

It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.

The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.