$\widetilde{(M \otimes_A N)} \cong \tilde{M} \otimes_{\mathcal O_X} \tilde{N}$ Let $A$ be a ring, $X = \textrm{Spec } A$, and $M$ an $A$-module.  For $U$ open, define $\tilde{M}(U)$ to be the abelian group consisting of all $m_{\mathfrak p} \in \prod\limits_{\mathfrak p } M_{\mathfrak p}$ such that locally, $m_{\mathfrak p} = \frac{m}{a}$ for some $m \in M, a \in A$.  Then $\tilde{M}$ is a sheaf of $\mathcal O_X$-modules.  If $N$ is another $A$-module, then $\tilde{M} \otimes_{\mathcal O_X} \tilde{N}$ is the sheaf associated to the presheaf 
$$U \mapsto \tilde{M}(U) \otimes_{\mathcal O_X(U)} \tilde{N}(U)$$
which is also a sheaf of $\mathcal O_X$-modules.  I'm trying to understand why 
$$\widetilde{(M \otimes_A N)} \cong \tilde{M} \otimes_{\mathcal O_X} \tilde{N}$$
as $\mathcal O_X$-modules.  This is II, 5.2(b) in Hartshorne.  All that is written for the proof is "$M \mapsto \tilde{M}$ commutes with tensor product, because this commutes with localization" which I don't understand.
I was trying to reason as follows, but I am not sure this is on the right track.  For $U$ open in $X$, let $S(U) = \{ a \in A : a \not\in \mathfrak p \textrm{ for all } \mathfrak p \in U \}$, which is multiplicatively closed.  Then I believe that $\mathcal O_X$ is the sheaf  associated to the presheaf of rings $\alpha: U \mapsto S(U)^{-1}A$, and $\tilde M$ is that associated to the presheaf of abelian groups $\beta: U \mapsto S(U)^{-1}M$.  
Then $\beta$ is a presheaf of $\alpha$-modules, which induces $\tilde M$ as a sheaf of $\mathcal O_X$-modules.  The same for $\gamma: U \mapsto S(U)^{-1}N$ and $\tilde{N}$.  Now $\delta: U \mapsto \beta(U) \otimes_{\alpha(U)} \gamma(U)$ is a presheaf of abelian groups whose associated sheaf is probably $\tilde{M} \otimes_{\mathcal O_X} \tilde{N}$.  But at the same time,
$$\delta(U) = S(U)^{-1}M \otimes_{S(U)^{-1}A} S(U)^{-1}N \cong S(U)^{-1}(M \otimes_A N)$$ 
and the associated sheaf is $\tilde{M \otimes_A N}$.  
 A: Let us construct an isomorphism $\widetilde{M\otimes_AN}\simeq\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}$ using only universal properties of tensor product, sheafification and localization.
Denote the presheaf $U\mapsto \tilde{M}(U)\otimes_{\mathcal{O}_X(U)}\tilde{N}(U)$ by $\mathcal{F}$. Then we have a morphism of presheaves $\mathcal{F} \to \tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}$ and any morphism of presheaves $\mathcal{F}\to \widetilde{M\otimes_AN}$ uniquely factors through $\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}$.
Given $(s,t)\in \tilde{M}(U)\times\tilde{N}(U)$ and $\mathfrak{p}\in U$, there are opens $\mathfrak{p}\in V_s,V_t\subset U$ such that $s\mid_{V_s}=\frac{m}{f}$ with $m\in M$ and $f\notin \mathfrak{q}$ for every $\mathfrak{q}\in V_s$ and similarly $t\mid_{V_t}=\frac{n}{g}$. So $V=V_s\cap V_t$ is an open neighborhood of $\mathfrak{p}$ on which both $s$ and $t$ are fractions and  $f,g\notin \mathfrak{q}$ for every $\mathfrak{q}\in V$. Furthermore, on $V$ we have $s\mid_{V}=\frac{m}{f}=\frac{gm}{fg}$ and $t\mid_{V}=\frac{n}{g}=\frac{fn}{fg}$.
That is, given $(s,t)\in \tilde{M}(U)\times\tilde{N}(U)$ and $\mathfrak{p}\in U$, there is an open $\mathfrak{p}\in V\subset U$ such that $(s,t)\mid_V=(\frac{m}{f},\frac{n}{f})$ with $m\in M$, $n\in N$ and  $f\notin \mathfrak{q}$ for every $\mathfrak{q}\in V$. Note that the last part means that $V\subset D(f)$.
We want a map $\tilde{M}(U)\times\tilde{N}(U)\to \widetilde{M\otimes_AN}(U)$. Given $(s,t)\in \tilde{M}(U)\times\tilde{N}(U)$ let $\{V_i\}$ be an open covering of $U$ such that $(s,t)\mid_{V_i}=(\frac{m_i}{f_i},\frac{n_i}{f_i})$ as above. We set $(s,t)\mid_{V_i} \mapsto \frac{m_i}{f_i}\otimes\frac{n_i}{f_i}\in \widetilde{M\otimes_AN}(V_i)$.
Now if $V_i \cap V_j$ is not empty, since $(s,t)\mid_{V_i\cap V_j}=(\frac{m_i}{f_i},\frac{n_i}{f_i})\mid_{V_i\cap V_j}=(\frac{m_j}{f_j},\frac{n_j}{f_j})\mid_{V_i\cap V_j}$ so $\frac{m_i}{f_i}\mid_{V_i\cap V_j}=\frac{m_j}{f_j}\mid_{V_i\cap V_j}$, there is a non empty open $W_{ij}\subset V_i \cap V_j$ and $g_{ij}\in A$ such that $g_{ij}(f_jm_i-f_im_j)=0$ in $M$ and $g_{ij}\notin \mathfrak{p}$ for every $\mathfrak{p}\in W_{ij}$. Similarly, there is $W'_{ij}\subset V_i \cap V_j$ and $g'_{ij}\in A$ such that $g'_{ij}(f_jn_i-f_in_j)=0$ in $N$ and $g'_{ij}\notin \mathfrak{p}$ for every $\mathfrak{p}\in W'_{ij}$. It follows that $$g_{ij}g'_{ij}(f_jm_i\otimes f_jn_i-f_im_j\otimes f_in_j)=g_{ij}g'_{ij}(f_jm_i\otimes f_jn_i-f_jm_i\otimes f_in_j+f_jm_i\otimes f_in_j-f_im_j\otimes f_in_j)=0$$ in $M\otimes_AN$. That is $$(\frac{m_i}{f_i}\otimes\frac{n_i}{f_i})\mid_{W_{ij}\cap W'_{ij}}=(\frac{m_j}{f_j}\otimes\frac{n_j}{f_j})\mid_{W_{ij}\cap W'_{ij}}\in \widetilde{M\otimes_AN}(W_{ij}\cap W'_{ij}).$$
We can cover $V_i\cap V_j$ by sets of the form $W_{ij}\cap W'_{ij}$ so $$(\frac{m_i}{f_i}\otimes\frac{n_i}{f_i})\mid_{W_{ij}\cap W'_{ij}}=(\frac{m_j}{f_j}\otimes\frac{n_j}{f_j})\mid_{W_{ij}\cap W'_{ij}}\in \widetilde{M\otimes_AN}(V_i\cap V_j).$$
Since $\widetilde{M\otimes_AN}$ is a sheaf the $ \frac{m_i}{f_i}\otimes\frac{n_i}{f_i}\in \widetilde{M\otimes_AN}(V_i)$ glue to a unique section over $U$, which we denote $s\otimes t$, and we obtain a map $\tilde{M}(U)\times\tilde{N}(U)\to \widetilde{M\otimes_AN}(U)$ for every open $U\subset X$.
The same sort of reasoning shows that these maps commute with the restrictions of the presheaves.
Next we show that the maps constructed above are bilinear.
Note that for $s,s'\in \tilde{M}(U)$ and $t\in\tilde{N}(U)$ we can find an open covering $\{V_i\}$ of $U$ such that $s\mid_{V_i}=\frac{m_i}{f_i}$, $s'\mid_{V_i}=\frac{m'_i}{f_i}$ and $t\mid_{V_i}=\frac{n_i}{f_i}$. It is clear that $$(s+s',t)\mid_{V_i}\mapsto (\frac{m_i}{f_i}+\frac{m'_i}{f_i})\otimes \frac{n_i}{f_i}=\frac{m_i}{f_i}\otimes \frac{n_i}{f_i}+\frac{m'_i}{f_i}\otimes \frac{n_i}{f_i}=s\otimes t\mid_{V_i}+s'\otimes t\mid_{V_i}$$ so $(s+s',t)\mapsto (s+s')\otimes t=s\otimes t+s'\otimes t$. The same goes for $(s,t+t')$. Similarly for $(\alpha s,t)$ with $\alpha\in \mathcal{O}_X(U)$, locally $\alpha$ is of the form $\frac{a}{g}$ so 
$$(\alpha s,t)\mapsto (\alpha s)\otimes t=\alpha (s\otimes t)=s\otimes (\alpha t)$$ and $(s,\alpha t)\mapsto s\otimes (\alpha t)$. Hence we have maps $\mathcal{F}(U)\to \widetilde{M\otimes_AN}(U)$ and since these maps commute with the restrictions we obtain a morphism of presheaves $\mathcal{F}\to \widetilde{M\otimes_AN}$.
So far we have constructed a morphism of sheaves $\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N} \to \widetilde{M\otimes_AN}$.
In order to show that this is an isomorphism we need to show that for every $\mathfrak{p}\in X$ the induced map of stalks $(\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N})_\mathfrak{p} \to (\widetilde{M\otimes_AN})_\mathfrak{p}$ is an isomorphism.
By the properties of sheafification $(\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N})_\mathfrak{p}=\mathcal{F}_\mathfrak{p}$ for every $\mathfrak{p}\in X$.
By the properties of the $\widetilde{\phantom{a}}$ construction we know that 
$$\widetilde{M}_\mathfrak{p}=M_\mathfrak{p} \quad \widetilde{N}_\mathfrak{p}=N_\mathfrak{p}\quad \widetilde{M\otimes_AN}_\mathfrak{p}=(M\otimes_AN)_\mathfrak{p}.$$
It is clear that $\mathcal{F}_\mathfrak{p}=M_\mathfrak{p}\otimes_{A_\mathfrak{p}}N_\mathfrak{p}$ and, by construction of the morphism, that the induced map $M_\mathfrak{p}\otimes_{A_\mathfrak{p}}N_\mathfrak{p} \to (M\otimes_AN)_\mathfrak{p}$ is given by $\frac{m}{f}\otimes\frac{n}{g}\mapsto \frac{m\otimes n}{fg}$ which is easily seem to be an isomorphism by the universal property of localization. 
