On a Riemannian manifold, it is well-known that two distinct non-closed geodesics can not meet tangentially, that is, if they meet at time $t_0$ and $t_1$, then they will meet at different angles. The proof uses the symmetric property of the Riemannian metric. As we do not have a symmetric property on a Finsler manifold, does the conclusion still hold? In other words, can we have a picture like the one below?
Two distinct non-closed geodesics in a Finsler manifold must be transversal to each other. So the situation you mentioned cannot appear, even in a non-symmetric Finsler manifold.
Suppose $\gamma$ is geodesic for a Finlser metric $F$ on a manifold $M$. Then $\bar \gamma$, i.e, $\gamma$ traversed in the reverse direction is a geodesic for the reverse Finsler metric $\bar F$ (see, for example, Section 2.5 of Ohta's book).
Thus, if $\gamma_1$ and $\gamma_2$ meets tangentially, you must have $\bar \gamma_1$ and $\bar\gamma_2$ are identical. But that means $\gamma_1$ and $\gamma_2$ are identical.