You made a typo in your computation of the stalks, to define $\gamma$: then numerator should be $\overline{n}$, not $\overline{n}\overline{p_i}^{k_i}$. About your specific questions :
1) By the sheaf property, a global section is the same as a family of sections on opens of a covering, as long as they glue nicely on intersections. Here you have three open (in fact clopen) points which are therefore disjoint : so a global section is just a triple of sections that glue nicely on empty intersections; you can see that this last condition is empty : a global section is just a triple of sections, one on each open point.
Now, if $x$ is an open point, a section on $\{x\}$ is the same as an element of the stalk, which you computed earlier.
2) Well the point is that each point ($(2), (3), (5)$) is open, because they're all closed (and there's finitely many of them). Therefore they form an open covering, and so we go back to what I explained in 1).
3) The point is again that the three points are open, and thus form an open covering. Since they are disjoint, the sheaf property says exactly that the product of the three restrictions is an isomorphism.
So I think the main point you missed was that as each of $(2),(3),(5)$ is closed, they are also open (this uses the fact that there's finitely many of them) and are pairwise disjoint, so the sheaf property applied to the open covering $\{\{(2)\}, \{(3)\}, \{(5)\}\}$ gives the isomorphism (together with your computation of the stalks)
This is not a robust answer, but provides a heuristic reasoning for why the approach proposed in the question may not work.
For a suitably arbitrary prime $p$, a pair of residues $x,y \pmod{p}$ are likely to be coprime i.e., their GCD is likely to be $1$ with more than $0.5$ probability.
For example, there are $13$ coprime pairs modulo $5$ (therefore, coprimality probability = $0.52$) and $93$ pairs modulo $13$ (coprimality probability = $0.55$) and the limit for the probability is $6 \over \pi^2$ as $p \rightarrow \infty$. This is approximately $0.61$. (See this question and answer for more info).
If the prime doesn't divide the integers $x,y$ then we have to ignore the residue pairs $(0, y)$ and $(x, 0)$ and then the effective coprimality probability would go up even higher.
So, using even a probabilistic argument, we are likely to see $1$ as the GCD of the residues quite often even when the numbers themselves are not coprime.
Using this and the pigeon-hole principle, we should be finding the actual GCD only by chance (and very low at that).
Best Answer
This is a situation where viewing the problem in greater generality makes things a lot less complicated. Suppose you have two commutative rings $R$ and $S$, and let $\varphi \colon R \to S$ be a (unital) ring homomorphism. Try to prove the following:
(You might recognize that you are actually demonstrating that the association $R \mapsto \operatorname{GL}_{n}(R)$ defines a functor from the category of commutative rings to the category of groups.)
Once you've proven these facts, you may deduce that $\operatorname{GL}_{n}(\varphi)$ is an isomorphism if $\varphi$ is an isomorphism. Your problem then doesn't require any specific require any specific knowledge, other than that the Chinese Remainder Theorem furnishes a ring isomorphism $\mathbb{Z}/N\mathbb{Z} \to \mathbb{Z}/p_{1}^{k_{1}}\mathbb{Z} \times \mathbb{Z}/p_{2}^{k_{2}}\mathbb{Z}$ defined as you suggest.