Say $0 \to G_1 \xrightarrow f G_2 \xrightarrow g G_3 \to 0$ is exact and gcd$(|G_1|=n, |G_3|=m)=1.$ Then the sequence splits.
We have that there exist $r,s \in \mathbb{Z}$ such that $mr+ns=1$. We then define a map $\beta:G_3 \to G_2$ such that $\beta(g_3)=mrg_2$ where $g_3=g(g_2)$.
I don't know how to show that this map is well defined.
I believe you want to say that $gcd(|G_1|=n,|G_3|=m)=1$. $G_1$ is a subgroup of $G_2$ thus by Lagrange, the cardinal of $G_1$ divides the cardinal of $G_2$.
You have $an+bm=1$, let $h:G_3\rightarrow G_2$ be a section. i.e $g(h(x))=x$, write $l(x)=h(x)^{an}$, we have $l(xy)l(y)^{-1}l(x)^{-1}=h(xy)^{an}h(y)^{-an}h(x)^{-an}=(h(xy)h(y)^{-1}h(x)^{-1})^{an}=1$ since the order of $G_1$ is $n$ and $h(xy)h(y)^{-1}h(x)^{-1}\in G_1$.
We also have $g(l(x))=g(h(x)^{an})=x^{an}=x^{1-bm}=x$ since the order of $G_3$ is $m$. We deduce that $l$ is a morphism of groups, thus the sequence splits.