I can't understand the red line well. Seemly, if $\tilde \gamma_1\cap \tilde \gamma_2$ has at least two points, it will contradict with the simple connectivity. I want to know why it is ?
PS(2024-3-26): As the hint of Moishe Kohan, the right way is to use the Hadamard Theorem.
Hadamard Theorem: Let $M$ be a complete Riemannian manifold, simply connected, with sectional curvature $K\le 0$. Then $M$ is diffeomorphic to $\mathbb R^n$, $n=\dim M$; more precisely $\exp_p:T_pM\rightarrow M$ is a diffeomorphism.
If there are at least two points, assuming two of them are $p,q$, then, there are two different vectors $u,v\in T_pM$ such that $\exp_pu=\exp_pv =q$. It contradict with $\exp_p$ is diffeomorphism.
Picture below is from the 260th page of do Carmo's Riemannian Geometry.
Suppose the intersection contains just one point, $P$. Where must that point go after the translation, i.e., what is $f(P)$? It must lie on both geodesics, because we're assuming both are (setwise) invariant under $f$. But the only point on both geodesics is $P$. So we get $f(P) = P$, i.e., $P$ is a fixed-point of $f$. But $f$ has no fixed points (by the opening clause of that sentence), so that's a contradiction.