The question is closely related to this one. Although the OP accepted already an answer there, I think that a subtely detail is still not solved and would like to ask about it here explicitely.
The OP asked:
Let $(X,\mathscr{O}_X)$ a ringed space and $\mathscr{F}$ an $\mathscr{O}_X$-module. Let $U\subseteq X$ be an open subset, and suppose we have $\mathscr{F}|_U=\mathscr{O}_U$, then for $x\in U$, how does this imply $\mathscr{F}_x=\mathscr{O}_{X,x}$? The explanation I saw was "because localization preserves exactness", but I don't quite understand how does it work here. Can somebody explain?
And I think that it might be considered as important part of this problem to understand not only why $\mathscr{F}_x=\mathscr{O}_{X,x}$ is here true as result, but by how explicitely a proof should work, which uses the fact that 'localization preserves exactness' as tool.
Some user's including myself posted a proof but no one used explicitely localization preserves exactness as an input. That is the question is if there is a proof of $\mathscr{F}_x=\mathscr{O}_{X,x}$ which uses this feature of localizations?
Question: "And I think that it might be considered as important part of this problem to understand not only why Fx=OX,x is here true as result, but by how explicitely a proof should work, which uses the fact that 'localization preserves exactness' as tool."
Answer: You must specify why you speak about "localization". If $i: U \rightarrow X$ and $j:\{x\}\rightarrow U$ is the inclusion maps it follows (since topological inverse image is a functor) that there is an isomorphism of functors
$$(i \circ j)^{-1}\cong j^{-1} \circ i^{-1},$$
hence for any sheaf of abelian groups $F$ on $X$ you get isomorphisms
$$ F_x \cong (i \circ j)^{-1}F \cong j^{-1}i^{-1}F \cong j^{-1}(F_U) \cong (F_U)_x$$
where $F_U$ is the restriction of $F$ to the open set $U$.
Note: Localization in commutative algebra is exact in the following sense: For any multiplicative subset $S \subseteq A$ where $A$ is a commutative unital ring and any short exact sqeuence
$$0 \rightarrow E \rightarrow F \rightarrow G \rightarrow 0$$
of $A$-modules, it follows the sequence
$$0 \rightarrow S^{-1}E \rightarrow S^{-1}F \rightarrow S^{-1}G \rightarrow 0$$
is an exact sequence of $S^{-1}A$-modules. But in your case you are speaking of ringed spaces and hence this method does not apply.
Since you are considering arbitrary ringed spaces you cannot "localize" - it has no meaning. Localization (in the sense of commutative algebra) applies when $(X, \mathcal{O}_X)$ is a scheme - it has an open cover $U_i:=Spec(A_i)$ of affine schemes.
If $f\in A$ where $A$ is a commutative unital ring and $f\notin \mathfrak{p}$ where $\mathfrak{p}$ is a prime ideal in A, let $S:=A-\mathfrak{p}$ and $T:=\{1,f,f^2,..,\}$. It follows $T \subseteq S$ is a subset.
There is a canonical map
$$i: A \rightarrow A_f:=T^{-1}A$$
and let $S_f:=i(S) \subseteq A_f$. There is a canonical isomorphism
$$(A_f)_{\mathfrak{p}}\cong S_f^{-1}T^{-1}A \cong S_f^{-1}(A_f) \cong S^{-1}A :=A_{\mathfrak{p}}$$
since localization is "transitive". Hence you get an isomorphism of rings
$$I1.\text{ }A_{\mathfrak{p}} \cong (A_f)_{\mathfrak{p}}.$$
If $X:=Spec(A), U:=D(f):=Spec(A_f)$ the isomorphism in $I1$ proves that there is an isomorphism of rings
$$\mathcal{O}_{X, \mathfrak{p}} \cong A_{\mathfrak{p}}\cong (A_f)_{\mathfrak{p}} \cong \mathcal{O}_{U, \mathfrak{p}}.$$
Hence for affine schemes a similar result holds using "transitivity" of localization - not "localization preserve exactness". Similarly if $E$ is any $A$-module and $\mathcal{E}:=\widetilde{E}$ it follows there is an isomorphism
$$\mathcal{E}_{\mathfrak{p}} \cong E_{\mathfrak{p}} \cong (E_f)_{\mathfrak{p}} \cong (\mathcal{E}_U)_{\mathfrak{p}}.$$
Hence for any quasi coherent $\mathcal{O}_X$-module you may use "transitivity of localization" to prove the same result.
Note: To take the stalk of a sheaf of abelian groups is a topological construction, and this makes sense for any sheaf of abelian groups on any topological space. In the case of an affine scheme $X:=Spec(A)$, it agrees with the $A_{\mathfrak{p}}$-module $E_{\mathfrak{p}}$ corresponding to the prime ideal $\mathfrak{p}$.
Question: "And yes, my understanding of 'localization' is exactly that one from commutative algebra you explaned in your Note. And the core of my question is if exactly this comm alg characterization can be explicitely used to prove the claim."
Answer: The answer is "no". If $(X, \mathcal{O})$ is an arbitrary locally ringed space you do not have an open cover $U_i:=Spec(A_i)$ of $X$ of affine schemes $U_i$. If you want to use localization in the sense of commutative algebra you need such an open cover and the relation $\mathcal{O}_x \cong A_{\mathfrak{p}_x}$ where $\mathfrak{p}_x \subseteq A$ is the prime ideal corresponding to $x\in U:=Spec(A) \subseteq X$.