Ytukyg

I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.

But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.

We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and

$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.

Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?

$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be

$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$

but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.

Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.

So if we take the exact sequence

$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$

where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.

That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are

$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$

So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.

$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.