(This is a cross-post of this unanswered math.stackexchange question)

In Edmond's 1982 paper Surface Symmetry II, at the bottom of page 145, he writes:

"Corollary - If $G$ is a split nonabelian metacyclic group acting freely on a surface $M$, then the natural map $\textbf{B} : \mathcal{FG}(G,M)^*\rightarrow H_2(G,\mathbb{Z})$ is surjective.

Proof: Since $H_2(G)\cong\Omega_2(BG)$, each element of $H_2(G)$ is represented by a free action on some surface, which can be chosen to be connected..."

Here, $M$ is a closed oriented surface, and $\mathcal{FG}(B,M)^*$ is the set of orientation-preserving equivariant homeomorphism classes of free actions of $G$ on $M$. The quotient $M/G$ does not depend on the particular chosen action, so let $N := M/G$. Then the set $\mathcal{FG}(B,M)^*$ seems to be the same as the set of surjective homomorphisms $\pi_1(N)\twoheadrightarrow G$ modulo $\text{Aut}(\pi_1(N))$.

The map $\textbf{B} : \mathcal{FG}(G,M)^*\rightarrow H_2(G,\mathbb{Z})$ is the one given by Eric Wofsey's answer to this question. Namely, every element $\eta\in\mathcal{FG}(G,M)^*$ gives $M$ the structure of a $G$-torsor over $N$, hence we get a map $N\rightarrow BG$, and the image of $\eta$ in $H_2(G,\mathbb{Z}) = H_2(BG,\mathbb{Z})$ is defined to be the image of the fundamental class of $N$ under the induced map on homology $H_2(N)\rightarrow H_2(BG)$.

My question is - Could someone help to elucidate the final quoted sentence? The author never explicitly defines what $\Omega_2(BG)$ is, though it seems that it is some kind of "(co?)bordism group", but there seem to be many variants on (co?)bordism groups, and without understanding exactly what he means, it's difficult to understand why the isomorphism $H_2(G)\cong\Omega_2(BG)$ would result in the map $\textbf{B}$ being surjective.

From another angle - is it possible to argue that $\textbf{B}$ is surjective without appealing to the isomorphism with the (co?)bordism group?

Is it possible to describe the map $\textbf{B}$ group-theoretically? Ie, given an $\text{Aut}(\pi_1(N))$-orbit of a surjection $\pi_1(N)\twoheadrightarrow G$, how do we produce an element of $H_2(G)$? Here, if $G$ has the presentation $$1\rightarrow R\rightarrow F_n\rightarrow G\rightarrow 1$$ where $F_n$ is a free group of rank $n$, then we know that $H_2(G) = (R\cap[F,F])/[F,R]$.