The following is Proposition 1.10 of A primer on mapping class group.
Let $\alpha$ and $\beta$ be two essential simple closed curves in a surface $S$. Then $\alpha$ is isotopic to $\beta$ if and only if $\alpha$ is (freely) homotopic to $\beta$.
My question is on the proof of this statement.
Proof. Suppose $\alpha$ and $\beta$ are homotopic and using bigon criterion, we may assume we can take an isotopy so that $\alpha$ and $\beta$ are disjoint. We assume $\chi(S)<0$ (Euler characteristic). Choose lifts $\tilde{\alpha}$ and $\tilde{\beta}$ of $\alpha$ and $\beta$ that have the same endpoints in $\partial\Bbb H^2$. There is a hyperbolic isometry $\phi$ that leaves $\alpha$ and $\beta$ invariant, and acts by translation on these lifts$^*$. As $\tilde{\alpha}$ and $\tilde{\beta}$ are disjoint, we may consider the region $R$ between them. The quotient $R' = R/\langle\phi\rangle$ is an annulus; indeed it is a surface with two boundary components with infinite cyclic fundamental group. A priori, the image $R''$ of $R$ in $S$ is a further quotient of $R'$. However, since the covering map $R'\to R''$ is single sheeted on the boundary, it follows that $R'\cong R''$. The annulus $R''$ between $\alpha$ and $\beta$ gives the desired isotopy. $\square$
I think the $^*$ part is very false. It says like $\alpha$ and $\beta$ are axis of $\phi$. $\tilde{\alpha}$ and $\tilde{\beta}$ are not geodesics, at least not both of them. To use $R/\langle\phi\rangle$, I think we should consider the unique geodesic representation $\gamma$ in the homotopy class of $[\alpha] = [\beta]$ so that we can choose a lift whose endpoints are the same endpoints of $\tilde{\alpha}$ and $\tilde{\beta}$ and axis of $\phi$. And I think we can homotope $\alpha,\beta$ and $\gamma$ (on the lifts) so that $\tilde{\gamma}$ is contained in $R$. Then I think the rest of them are fine we only need to change $R'\to R''$ is single sheeted on $\gamma$. Is it correct?
If $\tilde{\alpha}$ and $\tilde{\beta}$ are both geodesics, I agree that the marked assertion is false, as we would have an isometry with two axes. This is not the case (as I think you are aware). The picture is of one geodesic, $\tilde{\gamma}$, connecting the two endpoints $\partial \tilde{\alpha} = \partial \tilde{\beta} \in \partial \mathbb{H}^2$ and two squiggly curves which are "periodic along $\tilde{\gamma}$" in the sense that translation along $\tilde{\gamma}$ will preserve both $\tilde{\alpha}$ and $\tilde{\beta}$.
Level of confidence in the remainder of the answer: 80%. It appears to require a bit more argument than what's in the book, so I feel I'm missing something.
If $[\alpha] = [\beta] \in \pi_1(S)$ (after perhaps making $\alpha$ and $\beta$ intersect, and choosing basepoints), the isometry $\phi$ we're looking for is the deck transformation corresponding to $[\alpha] = [\beta]$. It's then clear that $\phi$ preserves the lifts of $\alpha$ and $\beta$. Unfortunately we only know they're freely homotopic, and therefore conjugate in $\pi_1(S)$. I believe this can be remedied with a "point-pushing isotopy" whereby we modify $\alpha$ ($=[\eta \beta \eta^{-1}])$ with the isotopy supported in a tubular neighborhood of $\eta$ ( homeomorphic to an annulus) found by dragging the center of the annulus along $\eta$. This modified $\overline{\alpha}$ will be equal to $[\beta] \in \pi_1(S)$.
I'm not enough of a real topologist to make that argument without being a little nervous, but hopefully it's helpful. If someone spots a mistake, please let me know.