Let $k$ be a field and $f:C\rightarrow B$ a morphism of smooth curves. Let $C'$, $B'$ and $f'$ be the extensions of $C$, $B$, and $f$ to $\overline{k}$.
Let $c\in C$, $b=f(c)$, $c'\in C'$ above $c$ and $b'=f'(c')$.
Is it true that the ramification index $e(c/b)$ is the same as the ramification index $e(c'/b')$?
As suggested by the author, I'm turning my comment into an answer.
It is true if $e(c/b)$ or $e(c′/b′)$ is equal to 1 and $k$ is perfect (thank you @Daniel Hast for detecting this), because then it's unramifiedness we're talking about, and it is detected by whether $\Omega^1$ vanishes. But $\Omega^1$ behaves nicely with base change (and it's a faithfully flat base change).
In general, I think it remains true if $k$ is perfect, because then the $C_{\overline{k}} \rightarrow C$ (same for $B$) correspond to unramified maps of valuation rings, thus $e(c/b)$ and $e(c′/b′)$ are both the ramification indices of $\mathcal{O}_{B,b}\rightarrow \mathcal{O}_{C′,c′}$ (using either $\mathcal{O}_{C,c}$ or $\mathcal{O}_{B′,b′}$ as intermediate rings).
So what I showed is that we can assume that $\overline{k}/k$ is purely inseparable. But then, this base change is a universal homeomorphism, and stranger things may happen (as shown in the answer linked by Daniel Hast again).