Let $f:Y\rightarrow X$ be a finite morphism of noetherian schemes. Let $\mathcal{F}$ be a coherent $\mathcal{O}_{Y}$-module. Then $f_{*}\mathcal{F}$ is a coherent $\mathcal{O}_{X}$-module.
Let $\{U_{i}\}$ be an open covering of $X$ with affines such that the $V_{i}=f^{-1}(U_{i})$ are affine open in $Y$. Let $\mathcal{F}|_{V_{i}}=\widetilde{M_{i}}$ with $M_{i}$ a finitely generated $\mathcal{O}_{Y}(V_{i})$-module. Then $(f_{*}\mathcal{F})|_{U_{i}} = \widetilde{M_{i}}$ where $M_{i}$ is now seen as an $\mathcal{O}_{X}(U_{i})$-module. It follows that $f_{*}\mathcal{F}$ is quasi-coherent. As $\mathcal{O}_{Y}(V_{i})$ is a finitely generated $\mathcal{O}_{X}(U_{i})$-module, and $M_{i}$ is a finitely generated $\mathcal{O}_{Y}(V_{i})$-module, we have that $M_{i}$ is a finitely generated $\mathcal{O}_{X}(U_{i})$-module. Thus $f_{*}\mathcal{F}$ is indeed coherent.
Question: Why do we have $(f_{*}\mathcal{F})|_{U_{i}}=\widetilde{M_{i}}$ as an $\mathcal{O}_{X}(U_{i})$-module?
Best Answer
This is more or less just unwinding the definition of the direct image.
First, reduce to the case that $X$ is affine so we don't have to keep writing $U_i$ and $V_i$ everywhere. Now we want to show that $f_*\mathcal{F}=\widetilde{M}$ as sheaves on $X$. By the definition of the direct image, we have that $(f_*\mathcal{F})(X)=M$, so we get a map $\widetilde{M}\to f_*\mathcal{F}$, and we want to show this is an isomorphism.
We do this in the only reasonable way: looking at stalks. We can calculate stalks on $X$ via affine opens of the form $D(g)$, and we see that $f^{-1}(D(g))=D(g)$ where on the right hand side we consider the image of $g$ in the coordinate ring of $Y$ under the pullback map. By the definition of $f_*\mathcal{F}$, we see that $(f_*\mathcal{F})(D(g))=\mathcal{F}(D(g))=M_g$. We also have $\widetilde{M}(D(g))=M_g$ by definition, and these identifications are compatible with the identification on global sections. So we get that the induced maps on stalks are isomorphisms and so $f_*\mathcal{F}=\widetilde{M}$.