Let $X=$ Spec$R$, $Y=$ Spec$S$, $f:X \to Y$ be a morphism of schemes. Let $M$ be a $R$-module, and let $\mathcal{F}=\tilde{M}$ be the sheaf on $X$ induced by $M$. How can I show that the pushforward of $\mathcal{F}$ by $f$, $f_* \mathcal{F}$ is the sheaf induced by $M_S$, where $M_S$ means $M$ considered as a $S$ module?
Here is my (feeble) attempt: Let $U$ be any open set of $Y$. $f_* \mathcal{F}(U)=\mathcal{F}(f^{-1}(U))=\{(\phi_p)_{p \in f^{-1}(U)}:\phi_p \in M_p $ and ...$\}$, meanwhile $\tilde{M_S}(U)=\{(\phi_q)_{q \in U}:\phi_q \in M_q $ and ...$\}$, ('...' contains some local conditions in the sheaf). Now the indexing $p \in f^{-1}(U)$ and $q \in U$ do not 'match up', by that I mean not all $q \in U$ is of form $f(p)$, where $p \in f^{-1}(U)$, so I do not see how one can construct a correspondence from $f_* \mathcal{F}(U)$ to $\tilde{M_S}(U)$. Also I do not recall any results relating the $M_p$ and $M_q$, hence I am having trouble seeing how the 2 sets can be considered equal.
I do believe I am missing something here, as a few algebraic geometry books I've flipped through state this as 'immediate from definitions'. Any help is appreciated!
This is a solution based on the hints of @Zhen Lin.
$$f_* \mathcal{F}(D(s))=\mathcal{F} (f^{-1}(D(s))) =\mathcal{F}(D(f^*s)) =M_{f^*s} =M_s =\tilde{M(D(s))}$$
Hence the 2 sheaves agree on the distinguished open sets $D(s)$, as these sets form a basis of the topology, we can use the gluing property of sheaves to see that they agree on every open set also.