I think your first paragraph is what I said. The choice of $\overline{\eta}$ gives me the choice of $\overline{s}$ (by normalizing $S$ in $\overline{\eta}$ and taking the residue field) which is why there is a canonical $\bar{x}\in X_{\bar{s}}(\overline{k})$ associated to $x$. But, what you didn't answer (unless I am misunderstanding you) is why this actually gives me a well-defined action of $\mathrm{Gal}(\overline{\eta}/\eta)$ on $(R\Psi\overline{\mathbb{Q}}_\ell)_{\overline{x}}$. My worry is that we should get the action just because the compatible action

Thank you :) In retrospect, that should have been pretty obvious!