Bipartite Graph Game
Let me spell out the solution.

We first show that Player 2 wins if there is a perfect matching $\pi$. Player 2's strategy is very simple: if the current path ends at a vertex $v$, she choose $\pi(v)$. Induction shows that $\pi(v)$ is never taken, since the edges chosen by Player 1 never belong to the perfect matching. Player 2 never gets stuck, and so wins.

The more complicated case is when there is no perfect matching. Let $\pi$ be a maximal matching, and let $x \in X$ be a vertex not in $\pi$. Since $x \notin \pi$, the edge $(x,y)$ chosen by Player 2 is not in $\pi$. If he can, Player 1 now choose $\pi(x)$, and induction again shows that Player 2 never chooses edges in $\pi$, and so Player 1 can choose an edge in $\pi$ as long as there is one adjacent to the vertex that Player 2 chose. If Player 1 gets stuck then the edges chosen by the players trace out an alternating path with Player 2 having one extra edge, showing that $\pi$ wasn't a maximal matching.