Unfortunately, as far as I know, nobody really explains such things as they are considered to be "too elementary". The most basic, briefest and down-to-earth reference I know for the needed background in Fuchsian groups is S.Katok's book "Fuchsian groups." Here are some proofs:

A Fuchsian group is called *elementary* if its limit set consists of at most 2 points; a group is called nonelementary otherwise. (E.g., every Fuchsian group $\Gamma$ with $Area(H^2/\Gamma)<\infty$ is nonelementary.) You also need to know that a nonelementary group contains at least two hyperbolic elements $\gamma_1, \gamma_2$ with disjoint fixed-point sets. (All this should be in Katok's book.)

Suppose now that $\Gamma$ is nonelementary Fuchsian and $g\in G:=PSL(2,R)$ belongs to the centralizer of $\Gamma$. Then $g$ would have to fix the fixed points of $\gamma_1$ and of $\gamma_2$, so it fixes four distinct points on the unit circle. Thus, $g=1$. Suppose now that $g\in N(\Gamma)$, the normalizer of $\Gamma$ in $G$. Then, unless $g=1$, the automorphism of $\Gamma$ induced by conjugation via $g$, is nontrivial. Thus, we obtain an injection $N(\Gamma)\to Aut(\Gamma)$. In particular, $g\in N(\Gamma)$ induces an inner automorphism of $\Gamma$ if and only if $g\in \Gamma$. Now, if $g\in G$ satisfies $g\Gamma g^{-1}\subset \Gamma$, then $g$ projects to an endomorphism $[g]$ of $S=H^2/\Gamma$ (just check that the above inclusion forces $g$ to preserve $\Gamma$-equivalence relation on $H^2$). If $g\in N(\Gamma)$ then the same argument shows that $[g]$ is an automorphism since $[g^{-1}]$ is its inverse. Clearly, $N(\Gamma)\to Aut(S), g\to [g]$ is a homomorphism whose kernel contains $\Gamma$. It is also immediate that $\Gamma$ is the kernel of this homomorphism. Thus, you get an embedding $N(\Gamma)/\Gamma \to Aut(S)$.

There are many sources for orbifolds (start with wikipedia article), but the key is that in the context you are interested in, orbifolds provide the "correct" generalization of the covering theory to the case of Fuchsian groups which do not act freely on $H^2$. For instance, you probably know that if $p: X\to Y$ is the universal cover (in the standard sense), then every homeomorphism $f: Y\to Y$ lifts to a homeomorphism $X\to X$ which normalizes the deck-group. The same works for orbifolds. Thus, if you treat $S$ as an orbifold $O$, then every (say, conformal) automorphism $f$ of $O$ lifts to a conformal automorphism $\tilde{f}$ of $H^2$ (the universal cover of $O$) which normalizes $\Gamma$. Hence, $\tilde{f}\in PSL(2,R)$ and, thus, belongs to $N(\Gamma)$. This establishes that the homomorphism
$$
N(\Gamma)/\Gamma \to Aut(O)
$$
is also surjective. (The fact that for $g\in N(\Gamma)$, $[g]$ is an automorphism of $O$, is immediate once you understand the orbifold definitions.)