Suppose we have two distinct $\gamma_1$ and $\gamma_2$, non-intersecting geodesic lines with disjoint endpoints on the hyperbolic plane. Assuming $\gamma_i$ cuts the hyperbolic plane into two half-planes $A_i$, $B_i$ satisfying $A_1 \subset A_2$ and $B_2 \subset B_1$.
On the disc model this just looks like the disc with two geodesics which don't intersect, with endpoints on the boundary, I think.
There is a Mobius map $f$ with $f(B_1) = B_2$
How do we show this map exists and it is necessarily hyperbolic?
So far, I have argued we can send the endpoints of one geodesic to the endpoints of the other in a nice way (so they don't flip) and since Mobius maps preserve orientation, we get $f(B_1) = B_2$ but I don't know if this is sufficient.
As for showing map is hyperbolic, I don't know where to begin. I think we have to use the fact that a map from the interval to itself has a fixed point
Take a third geodesic $\eta$ that is perpendicular to both $\gamma_1$ and $\gamma_2$, then use a hyperbolic translation along it. You can visualize this in the upper half space model by arranging $\gamma_1$ and $\gamma_2$ as concentric circles centered on the origin so that $\eta$ be the imaginary axis, and then finding $a$ such that the Möbius transformation $z\mapsto az$ takes one of the $\gamma_i$ to the other.
(You can see that the geodesic $\eta$ exists by parametrizing one of the $\gamma_1$ by some parameter $t$, considering all perpendiculars $\eta_t$ and the angles they make with $\gamma_2$, and applying the intermediate value theorem.)
Here's another argument that doesn't rely on working in the upper half plane. You can also explicitly construct the perpendicular in the disc model. Use one isometry to make one of the $\gamma_i$ coincide with the real axis. Then use a hyperbolic translation along the real axis to arrange the other geodesic so it is centered on the imaginary axis. Now the geodesic coinciding with the imaginary axis is perpendicular to both of them.