The statement is not true. The universal covering of $T^2$ is homeomorphic to $\mathbb R^2$, and $\mathbb R^2$ is homeomorphic to $\mathbb H^2$, so there exists a universal covering map $\mathbb H^2 \to T^2$. But $T^2$ does not have a hyperbolic metric, by the Gauss-Bonnet theorem.
The correct statement is that if there exists a covering map $f : \mathbb H^2 \to S$ such that the deck transformation action of $\pi_1(S)$ on $\mathbb H^2$ is an action by isometries of $\mathbb H^2$ then $S$ admits a Riemannian metric of constant negative curvature.
The proof is to take any open subset $U \subset S$ which is evenly covered by $f$, choose an open subset $\tilde U \subset \mathbb H^2$ such that $f$ maps $\tilde U$ to $U$ by a homeomorphism, and define the Riemannian metric on $U$ as the pushforward via the map $f$ of the Riemannian metric on $\tilde U$. The hypothesis that the deck action is by isometries is needed here to show that the Riemannian metric on $U$ is well-defined independent of the choice of $\tilde U$.
The converse is only true with an additional hypothesis as well, namely that the Riemannian metric on $S$ is geodesically complete. The proof takes some work, but you can find it carefully written up in textbooks on hyperbolic geometry.