stalk of a direct image sheaf under a finite morphism Let $f: X \rightarrow Y$ be a finite surjective morphism of schemes, and $\mathscr{F}$ a coherent sheaf of $\mathscr{O}_X$-modules on $X$. I find it confusing to understand the stalk $(f_* \mathscr{F})_y$ at $y \in Y$ in terms of the finite number of stalks $\mathscr{F}_x$ where $f(x) = y$. A section of $(f_* \mathscr{F})_y$ can be restricted to be in $\mathscr{F}_x$, so if $f^{-1}(y) = \{ x_1, ..., x_n \}$, I see that
$$(f_* \mathscr{F})_y \longrightarrow \underset{i = 1, ..., n}{\oplus} \mathscr{F}_{x_i},$$
but is there a better way to understand the structure of $(f_* \mathscr{F})_y$ in terms of all $\mathscr{F}_{x_i}$?
 A: I think the scheme theory might be obscuring things here, finite morphisms are affine, so we should translate this to rings to see whats happening. Explicitly, pick an open affine $U$ containing $y$, then its preimage $f^{-1}(U)$ is also affine, and $f^{-1}(U)\rightarrow U$ is also finite.
I'll work with the sheaves being $O_X$ and $O_Y$, since the general case is just what happens to modules over these rings, and the localisation/stalk considerations result from tensoring with these rings.
So applying the equivalence between affine schemes and rings here, we have a (module finite) ring morphism $R\rightarrow S$. The stalk at $y$ corresponds to the localisation of $R$ at some prime $\mathfrak{p}$, and the stalks at the $x_i$ are the localisations of $S$ at the primes $\mathfrak{q_i}$ lying over $\mathfrak{p}$. So the question is then a classic part of commutative algebra, what is the relation between the localisation of $S$ with respect to the ideal $\mathfrak{p}$ of $R$ and the localisations of $S$ with respect to the $\mathfrak{q_i}$?
We have two important maps, $R_\mathfrak{p}\rightarrow S_\mathfrak{p}$, the localisation of $R\rightarrow S$ with respect to $\mathfrak{p}$, which will still be finite, and the maps $S_\mathfrak{p}\rightarrow S_\mathfrak{q_i}$ coming from the fact that $\mathfrak{p}\subset\mathfrak{q_i}$. These second maps are the induced maps on stalks in our geometric picture. Taking the product of these, we have $S_\mathfrak{p}\rightarrow \prod_i S_{\mathfrak{q_i}}$, thats your map. However, these local maps $S_\mathfrak{p}\rightarrow S_\mathfrak{q_i}$ aren't the easiest to actually do stuff with, for instance, they generally aren't module finite. 
Thus far, none of our maps have been isomorphisms either, and this isn't too surprising, all of our local rings are localisations of $R$ and $S$, and so they still see global behaviour, by taking the field of fractions for instance, if they were a domain. One way to remedy this is to take completions with respect to the ideals, an operation that goes "more local" than localisation. If you do this, you do get a module finite ring extension $\widehat{R}_\mathfrak{p}\rightarrow \widehat{S}_\mathfrak{q_i}$, and in some cases that map you have becomes an isomorphism once completed, for instance in a finite morphism of dedekind domains. In the dedekind domain setting this "ultralocal" approach is incredibly useful, as it lets you use linear algebra (discriminants, etc) locally, which isn't possible without module finiteness, thus requiring completion.
Number theory is a good place to see these concepts in action, eg, working out the stalks over $(2),(3),(5)$ in finite morphism $Spec(\mathbb{Z}[i])\rightarrow Spec(\mathbb{Z})$, or doing the same for a finite morphism of curves if you want to be more geometric.
