As Jonathan said, the answer is always, and the left adjoint $f_!$ has a simple description as the "extension by $\emptyset$" functor. Think of an open immersion $j:U\rightarrow X$ of topological spaces, then we have the usual extension by $\emptyset$ functor by sheafififying the presheaf

$j_!\mathcal{F}(W)=\mathcal{F}(W)$ if $W\subset U$

$j_!\mathcal{F}(W)=\emptyset$ otherwise.

It's an easy calculation to show that this is left adjoint to $j^*$.

Translating this to the étale topology, if $f:X\rightarrow Y$ is an étale morphism of schemes, we have the extension by $\emptyset$ functor by sheafififying the presheaf

$f_!\mathcal{F}(W\overset{f}{\rightarrow} Y)=\coprod_{g\in\mathrm{Hom}_Y(W,X)}\mathcal{F}(W\overset{g}{\rightarrow} X)$

and again it's fairly easy to verify that $f_!$ does what we want it to.

It's worth noting that to get an adjoint for abelian sheaves, we need to replace this coproduct by the coproduct in the category of abelian sheaves, i.e. direct sum. Thus $f_!\mathcal{F}$ depends on whether we are considering $\mathcal{F}$ as a sheaf of sets or of abelian groups.

It's possibly also worth noting that we don't get a geometric morphism $f^*:\mathrm{Sh}(Y)\leftrightarrows \mathrm{Sh}(X):f_!$ because $f_!$ doesn't preserve limits. Despite this, the version of $f_!$ for abelian sheaves is actually exact!