It is enough to prove this in the case where $Y$ is $Spec$ of a strictly Henselian ring. I think, one sees the main point in the argument already in the following special case:

Let $Y=Spec(K)$ and $X=Spec(E)$ where $E/K$ is a finite separable extension of fields. Denote $\pi: X\to Y$ the canonical map. Let $\overline{E}$ be an algebraic closure of $E$ and $\overline{x}$ (resp. $\overline{y}$) be the corresponding geometric point of $X$ (resp $Y$). Let $L/K$ be a finite separable extension contained in $\overline{E}$ and containing the Galois closure of $E$. If $F$ is a sheaf on $X$, then $\pi_*F(Spec(L))=F(Spec(L\otimes_K E))$ and $Spec(L\otimes_K E)= \coprod_{i=1}^d Spec(L)$, where $d=[E:K]$. Hence $\pi_* F(Spec(L))=F(Spec(L))^d$. Now take an inductive limit over such $L$, in order to obtain $\pi_*F_{\overline{y}}=F_{\overline{x}}$.

Hope this is of some use...

It is enough to prove this in the case where $Y$ is $Spec$ of a strictly Henselian ring. I think, one sees the main point in the argument already in the following special case:

Let $Y=Spec(K)$ and $X=Spec(E)$ where $E/K$ is a finite separable extension of fields. Denote $\pi: X\to Y$ the canonical map. Let $\overline{E}$ be an algebraic closure of $E$ and $\overline{x}$ (resp. $\overline{y}$) be the corresponding geometric point of $X$ (resp $Y$). Let $L/K$ be a finite separable extension contained in $\overline{E}$ and containing the Galois closure of $E$. If $F$ is a sheaf on $X$, then $\pi_*F(Spec(L))=F(Spec(L\otimes_K E))$ and $Spec(L\otimes_K E)= \coprod_{i=1}^d Spec(L)$, where $d=[E:K]$. Hence $\pi_* F(Spec(L))=F(Spec(L))^d$. Now take an inductive limit over such $L$, in order to obtain $\pi_*F_{\overline{y}}=F_{\overline{x}}^d$.

Hope this is of some use...