Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

The better way to define a homomorphism from $\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/\gcd(m,n)\mathbb{Z}$ is via the universal property.

Note that the map $\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z}$ defined by $$(x+m\mathbb{Z},y+n\mathbb{Z})\mapsto xy+\gcd(m,n)\mathbb{Z}$$ is well-defined and also bi-linear, thus by the universal property of tensor product, there is a linear map $f:\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z}$ such that $$f(x+m\mathbb{Z}\otimes y+n\mathbb{Z})=xy+\gcd(m,n)\mathbb{Z}.$$ Verify that the linear map $g:\mathbb{Z}/\gcd(m,n)\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z}$ defined by $$g(z+\gcd(m,n)\mathbb{Z})=(z+m\mathbb{Z})\otimes (1+n\mathbb{Z})$$ is well-defined, and we also have $g\circ f=1, f\circ g=1$, thus $f$ is isomprhism. To see $g$ is well-defined, you may use the equality that $\gcd(m,n)=am+bn$ for some integers $a,b\in\mathbb{Z}$.

The part that is missing is pretty much the essence of the following (incomplete) alternative proof.

Determine the kernel of $$ \begin{array}{rlrl} g: & \mathbb{Z} & \rightarrow & (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \\ & z & \to & z (1 \otimes 1) \end{array} $$ That is: When is it true that $z (1 \otimes 1)$ is null? Since it is true for $z \in m\mathbb{Z} \cup n\mathbb{Z}$, then you know that it is true for the ideal generated by it: $\langle \mathrm{gcd}(m,n)\rangle \subset \mathrm{ker}(g)$.

You know that the map is surjective because $1 \otimes 1$ is a generator. If you show that $\mathrm{ker}(g) \subset \langle \mathrm{gcd}(m,n) \rangle$, you will have the isomorphism you claim. This is the part you are missing. It is equivalent to showing that your $f$ is well-defined.

So, the conclusion is that it is not right.