First of all, you need to know that your surface $S$ admits a hyperbolic structure, i.e. is homeomorphic to the quotient $H^2/\Gamma$ of the hyperbolic plane $H^2$ by a group of isometries $\Gamma$ acting properly discontinuously and freely on $H^2$. See e.g. my answer here, although if you do not fix a conformal structure in advance, constructing a hyperbolic structure is easier, see for instance, here. In the same lectures by Aramayona you will find all the needed background material for what is written below.
Every oriented homotopically nontrivial loop $\alpha$ on $S$ represents a (unique up to conjugation) element $\gamma\in\Gamma$. The latter acts on $H^2$ as a hyperbolic isometry: It has exactly two fixed points on the boundary circle $S^1$ of $H^2$. The loop $\alpha$ admits a lift $\tilde{\alpha}$ which is invariant under $\gamma$. The lift $\tilde{\alpha}$ has exactly two limit points in the boundary circle $S^1$: These are the fixed points of $\gamma$. I will use the notation $Fix(\gamma)$ for the fixed-point set of $\gamma$ in $S^1$.
Conversely, if $\gamma$ preserves a (not necessarily simple) parameterized arc $\tilde\beta\subset H^2$, then the limit (accumulation) points of $\tilde\beta$ in $S^1$ are exactly the two fixed points of $\gamma$. Therefore, for each $\gamma'\in \Gamma$, if $\gamma'(\tilde\alpha)$ is $\gamma$-invariant, then $\gamma'$ preserves the fixed-point set $Fix(\gamma)\subset S^1$: It either fixes $Fix(\gamma)$ pointwise or swaps the two fixed points. The latter cannot happen since $\Gamma$ is torsion-free.
Lemma. The stabilizer $\Delta$ of $Fix(\gamma)=\{x_1, x_2\}$ in $\Gamma$ is an infinite cyclic group.
Proof. Let $L\subset H^2$ be the hyperbolic geodesic with the limit points $x_1, x_2$. Since there is a unique such geodesic, the line $L$ has to be invariant
under $\Delta$. Therefore, $\Delta$ acts freely as a properly discontinuous group of isometries of $L\cong {\mathbb R}$. The isometry group of ${\mathbb R}$ is the semidirect product of ${\mathbb R}$ (acting via translations) and ${\mathbb Z}_2$ (whose generator is the reflection $t\mapsto -t$). I will leave it to you as an exercise to show that every discrete infinite subgroup of ${\mathbb R}$ is isomorphic to ${\mathbb Z}$. Hence, $\Delta$ is isomorphic to ${\mathbb Z}$. qed
Therefore, every element $\gamma'\in \Gamma$ which sends $\tilde\alpha$ to a $\gamma$-invariant arc, belongs to the cyclic group $\Delta$. The latter contains $\langle \gamma\rangle$ as a finite index subgroup (since $\gamma\ne 1$). Therefore, the preimage of $\alpha$ in $H^2$ contains only finitely many $\gamma$-invariant arcs.
