Your space has a smooth $k'$-rational point $x$ with $k'/k$ finite. In the neighborhood of this point, the space is isomorphic to a disc over $k'$. On such a disc, it is not difficult to find a $k''$-rational point with $k''$ not linearly disjoint from $L$ over $k$ (at least if $k$ is not trivially valued and if $L \ne k$). This implies that there are several points of $X_L$ over $x$, hence the cover is not trivial.

In your particular case, $X_L$ has a point, so it is isomorphic to $P^{1,\mathrm{an}}_L$, hence simply connected. If your covering $X_L \to X$ were a covering, it would then be a universal covering. But we know that Berkovich curves retract by deformation onto graphs, so the topological fundamental group of $X$ is a free group. In particular, the universal covering of $X$ is either $X$ itself or of infinite degree, and we get a contradiction.