Given an affine scheme $X = \text{Spec}(A)$, then from an $A$-module $M$ we can form the associated sheaf $\tilde{M}$, where $$ \tilde{M}(D_f) = M_f = M \otimes_A A_f. $$
Now, given $A$-modules $M$ and $N$, we can compare $\tilde{N} \otimes_{\mathcal{O}_X} \tilde{M}$, and $(N \otimes_A M)^\tilde{}$.
First, $$ (N \otimes_A M)^\tilde{}(D_f) = (N \otimes_A M)_f = N_f \otimes_{A_f} M_f. $$
Does this extend to general open $U$? I.e. is
$$ (N \otimes_A M)^\tilde{}(U) = N(U) \otimes_{\mathcal{O}_X(U)} M(U) $$ true?
For example this is written in Algebraic Geometry and Arithmetic Curves, Prop. 5.1.12 by Qing Liu, but I cannot see why. I find this surprising because I am under the impression that $$ U \mapsto N(U) \otimes_{\mathcal{O}_X(U)} M(U), $$ is only a presheaf in general. Even if a presheaf agrees with a sheaf on a basis, as it does here, it is not necessarily a sheaf.
If this is true, this says that $\tilde{N} \otimes_{\mathcal{O}_X} \tilde{M} = (N \otimes_A M)^\tilde{}$.
If this is not true, then a counterexample would be appreciated.
First, the sheaf associated to $M\otimes_A N$ is $\widetilde{M}\otimes_{\mathcal{O}_X} \widetilde{N}$ without doing any of the calculations you worry about: see Stacks 01I8, Hartshorne proposition II.5.2, etc.
Next, your citation of Liu's text is incomplete - my copy of the text only makes that claim for affine open $U\subset X$. Indeed, $(\mathcal{F}\otimes\mathcal{G})(U)\neq \mathcal{F}(U)\otimes_{\mathcal{O}_X(U)} \mathcal{G}(U)$ in general, as the following example shows:
Let $X=\Bbb A^3_k$, let $\mathcal{F}=(k[x,y,z]/(x))^\sim$, let $\mathcal{G}=(k[x,y,z]/(y))^\sim$, and take $U$ to be the complement of the origin. Then $\mathcal{F}(U)=k[x,y,z]/(x)$, $\mathcal{G}(U)=k[x,y,z]/(y)$, $\mathcal{O}_X(U)=k[x,y,z]$, giving $\mathcal{F}(U)\otimes_{\mathcal{O}_X(U)}\mathcal{G}(U) = k[x,y,z]/(x,y)\cong k[z]$, while $(\mathcal{F}\otimes\mathcal{G})(U)\cong (k[x,y,z]/(x,y))^\sim(U)$, which is $k[z]_z$.