Let $f:X\to Y$ be a finite cover of smooth algebraic varieties, branched along a divisor $R\subset Y$. Let $E$ be a vector bundle on $Y$. What is the relation between the chern classes of $E$ and the chern classes of $F:=f_*(f^*E)$?

$\begingroup$ $f_*(f^*E)=E\otimes f_*\mathcal{O}_X$. $\endgroup$ – Mohan Oct 23 '13 at 15:25
You need to know what is the splitting of $f_* \mathscr{O}_X$, which depends on the cover.
Let us consider the double cover case, which is the easiest one. Then $$f_* \mathscr{O}_X = \mathscr{O}_Y \oplus L^{1},$$ where $L$ is a line bundle on $Y$ such that $R \in 2L$. Then, given any vector bundle $E$ on $Y$, by using projection formula one obtains $$F=f_*(f^*E)=f_*(f^*E \otimes \mathscr{O}_X) = E \otimes f_* \mathscr{O}_X = E \oplus (E \otimes L^{1}),$$ so the computation of the Chern classes of $F$ is now a straightforward application of the splitting principle.
For general finite covers of degree $d$, one can only say that $$f_* \mathscr{O}_X = \mathscr{O}_Y \oplus V,$$ where $V$ is a vector bundle on $Y$ of rank $d1$. As before, one obtains $$F=f_*(f^*E)=f_*(f^*E \otimes \mathscr{O}_X) = E \otimes f_* \mathscr{O}_X = E \oplus (E \otimes V),$$ but now the computations of Chern classes will be more difficult since in general $V$ is indecomposable.
In some lucky cases, for instance when the cover $f \colon X \to Y$ is Galois with abelian Galois group, $V$ splits completely into line bundles and then, if one is able to identify $V$, the computation can be made again by using the splitting principle.
By the projection formula $f_*(F^*E)\cong E\otimes f_*\mathcal{O}_X$, so $\mathrm{ch}(f_*(f^*E))=\mathrm{ch}(E).\mathrm{ch}(f_*\mathcal{O}_X)$. By GrothendieckRiemannRoch $\mathrm{ch}(f_*\mathcal{O}_X)$ is the pushforward of the relative Todd class $\mathrm{Todd}_{X/Y}$, so in principle you can compute the Chern character, hence the Chern classes, of $f_*(F^*E)$.