I would say that your proof is not particularly common, certainly I've never seen it before (admittedly this doesn't mean much). The proof in Hartshorne is brief because it is one of those scenarios where doing the "obvious" thing works. Here is the proof that I think Hartshorne had in mind (which is entirely constructive):
Let $T$ be the tensor pre-sheaf of $\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}$ and $\theta : T \rightarrow \tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}$ be the canonical morphism. We may define the "obvious" morphism $\varphi: T \rightarrow \tilde{(M\otimes_A N)}$ that over each open set sends $f\otimes g$ to $f\cdot g$.
Explicitly, for each $P \in \operatorname{Spec}(A)$, $(f\cdot g)(P) :=f(P)\cdot g(P) \in (M\otimes_A N)_P$ where we are implicitly using the canonical isomorphism from $M_P\otimes_{A_P} N_P$ to $(M\otimes_A N)_P$. This is where we are using that localisation commutes with tensor products.
You should check that on the stalk at $P$, $\varphi_P$ gives the canonical isomorphism $M_P\otimes_{A_P} N_P \rightarrow (M\otimes_A N)_P$.
Finally, by the universal property of sheafification, $\varphi$ factors through some morphism $\bar\varphi:\tilde{M}\otimes_{\mathcal{O}_X}\tilde{N}\rightarrow \tilde{(M\otimes_A N)}$, i.e. $\varphi = \bar\varphi\circ\theta$ and since $\varphi$ is an isomorphism on stalks, $\bar\varphi$ is an isomorphism.
As frequently happens in Hartshorne, a lot of algebraic detail is left out,and you can see why. If he included all of the details like this every time this kind of problem arose, the book would be $2-3$ times as long!
For the construction Vakil/Wedhorn and Görtz use, this is simply using the fact that you know what the sections of $\mathcal{O}_X, \tilde{M}$ and $\tilde{N}$ look like over the distinguished open affine pieces $D(f)$ of $X$. Since these form a basis for the topology on $X$ you can define the morphism $\varphi$ explicitly on these open pieces as the isomorphism $M_f\otimes_{A_f} N_f\rightarrow (M\otimes_A N)_f$ (this is again using the fact that tensor product commutes with localisations) and then "glue" this to get a morphism defined on any open set, which is an isomorphism on stalks since it is an isomorphism on the sections of a basis for the topology. You then still need to produce $\bar\varphi$ in the same way.
Which proof you prefer is entirely a matter of choice, they're essentially the same. The second one is probably cleaner, in some sense. The actual morphism looks nicer on each affine piece, but the price you pay is having to work slightly harder to show that the morphisms commute with restriction, depending on how happy you are to say it is "obvious".
