Suppose $S \cong \mathbb{H^2}/\Delta$ where $\Delta$ is a discrete subgroup of $PSL(2, \mathbb{R})$
I am told that $\Delta \subset PSL(2, \mathbb{R})$ is canonically isomorphic to $\pi_1(S)$.
I am probably missing a lot of theory, as I don't understand why this is. Any explanation would be greatly appreciated!
First, this is not true universally. For this to be true one must require that $\Delta$ is not only discrete, but that it acts freely, meaning that each nontrivial element of $\Delta$ has no fixed points in $\mathbb{H}^2$.
But second, as long as $\Delta$ is discrete and free, then this follows from the theory of covering spaces: the quotient map $f : \mathbb{H}^2 \to \mathbb{H}^2 / \Delta = S$ is a covering map, the domain $\mathbb{H}^2$ is simply connected so $f$ is a universal covering map, and the fundamental theorem of universal covering spaces has exactly the conclusion you want, namely that the group $\Delta$, which is the "deck transformation group" of the map $f$, is isomorphic to $\pi_1(S)$.