Suppose $f:X \to Y$ is a finite morphism, lets say of surfaces for simplicity.
If $|A|$ is a very ample linear system on $X$ then any general $H \in |A|$ is smooth, in any characteristic, assuming say your base field is algebraically closed. Now in characteristic zero a general element of $f^{*}|A|$ is smooth also by Bertini, and while a general element of $|A|$ need not pullback to a general element of $f^{*}|A|$ it seems to be well-accepted (or at least well-used) that the pullback of a general $H$ is smooth. This second part is typically said to be true by Bertini's theorem, however I wonder if Bertini in full generality is really needed, or even what is really going on.
Suppose for example that f is separable and tamely ramified, perhaps say of degree $d$ less than the characteristic. Is the pullback of a general $H$ smooth? I think we can argue it is smooth away from the ramified locus, so it is necessary to show we don't get self intersection over ramification locus, but it is not clear to me how one would do so, in any characteristic.
I've seen results along the lines of, suppose $X \to \mathbb{P}^{n}$ induces seperable residue field extensions over every point, then the pullback of a general hypersurface $H$ to $X$ is smooth. Which I suppose would suffice in characteristic zero. It is unclear to me whether such an $f$ as I've given above would satisfy this hypothesis but I suspect not.