It is a theorem that for a group $G$, all Eilenberg–Mac Lane spaces for $G$ are homotopy equivalent. (Section 1.B in Hatcher's Algebraic Topology is my reference for Eilenberg–Mac Lane spaces; this fact is proved there.)
In fact, something stronger is true. If $X$ and $Y$ are Eilenberg–Mac Lane spaces for $G$ and $x\in X$ and $y\in Y$ are points, any isomorphism $\pi_1(X,x)\to \pi_1(Y,y)$ is induced by a a based map $h\colon(X,x)\to(Y,y)$ that is a homotopy equivalence and unique up to a homotopy fixing $x$. (This is a restatement of 1.B.9 in Hatcher.)
So, having shown that $\Gamma\backslash H$ and $\Gamma\backslash V$ are Eilenberg–Mac Lane spaces, all it remains to show is that $f$ induces an isomorphism on fundamental groups. I'd prefer to leave that step to you, although I'm happy to say more about it if you'd like. This completes the proof because the proposition I quoted guarantees that the inverse isomorphism is induced by a map that is a homotopy inverse for $f$.
Since we agree that both spaces are Eilenberg–Mac Lane spaces, let's show that $f$ induces an isomorphism on fundamental groups. Firstly, we know that the action of $\Gamma$ on $H$ is a covering space action, so given a lift of the basepoint, say $1_H \in H$ and its image in $\Gamma\backslash H$, elements of the fundamental group of $\Gamma\backslash H$ are in one-to-one correspondence between homotopy classes of paths from $1_H$ to $g.1_H = g \in H$ for $g \in \Gamma$. Fixing $g$, let $\gamma_g \colon [0,1] \to H$ be such a path. Because $\tilde f\colon H \to V$ is continuous, $\tilde f\circ \gamma_g$ is a path from $\tilde f(\gamma_g(0))=\tilde f(1_H) = 0\in V$ to $\tilde f(\gamma_g(1)) = \tilde f(g) = g(0)$. But this homotopy class of paths is exactly those that correspond to $g$ as an element of the fundamental group of $\Gamma\backslash V$! Is this helpful?