If I've got the right idea of how the group operation of a semi-direct product works, then for some $c\in \mathrm{Hom}(\mathbb{Z}/n\mathbb{Z},\mathrm{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right))$ and $([x_1],[y_1]), ([x_2],[y_2]) \in \mathbb{Z}/m\mathbb{Z}\ltimes\mathbb{Z}/n\mathbb{Z}$,
\begin{align*} ([x_1],[y_1])\cdot([x_2],[y_2]) &= \left([x_1][x_2],c([x_2])[y_1][y_2]\right)\\ &= \left([x_1x_2],c([x_2])[y_1y_2]\right) \end{align*}
and I want to show that this commutes if and only if $\gcd(m,\varphi(n)) = 1$. I've shown that $\mathrm{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right)\cong \left(\mathbb{Z}/n\mathbb{Z}\right)^\times$, but I'm not sure where to go from here. Any assistance would be appreciated.
First of all, as anon commented, this question is ill defined right now: The structure of $\mathbb{Z}/m\mathbb{Z} \ltimes_c \mathbb{Z}/n\mathbb{Z}$ depends on the choice of $c:\mathbb{Z}/m\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/n\mathbb{Z})$, hence the $c$-subscript I've written. It's a true statement that every such semidirect product is abelian if and only if $m$ and $\varphi(n)$ are coprime.
Secondly, you don't have the composition law right. Writing $x \cdot y$ for the action of $x \in \mathbb{Z}/m\mathbb{Z}$ on $y \in \mathbb{Z}/n \mathbb{Z}$, we want $$(x_1,y_1)(x_2,y_2) = (x_1 x_2, (x_2 \cdot y_1) y_2).$$ It looks like you've written $$(x_1 x_2, x_2 \cdot (y_1 y_2)).$$
Ok, now let's prove our statement. It's clear from this composition law that if every $x$ acts as the identity on every $y$, i.e, if $c$ is the trivial map, then our semidirect product is actually a direct product, so our group is abelian. The image of $c$ must be a subgroup of $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$, but this group has size $\varphi(n)$, so by Lagrange's Theorem the size of the image divides $\varphi(n)$. The size of the image also must divide $n$ since it can be identified with a quotient of $\mathbb{Z}/m \mathbb{Z}$ by the first isomorphism theorem, so what does this mean about the image when $m$ and $\varphi(n)$ are coprime?
Conversely, if there is some common divisor, $p$, then we get a nontrivial map $c$ by sending $1$ to $\sigma$, where $\sigma$ is some automorphism of order $p$. Then there is some $x$ and $y$ with $x \cdot y \neq y$. Then $(1,y)(x,1) = (x,y \cdot x)$, but $(x,1)(1,y) = (x,y)$, so the group is not abelian.