I'm answering the edit: what is the action of $G=\pi_1(S)$ on $G'/G''$? A first thing to notice is that $G'$ acts trivially on $G'/G''$, so the action factors through to an action of $G/G'$ on $G'/G''$. The $\mathbb{Z}[G/G']$-module $G'/G''$ is known as the *first Alexander module* of $G$. (If $G$ is the fundamental group of $S^3-K$ for a knot, then this module is where the Alexander polynomial comes from.)

Topologically, $G/G'=H_1(S)$ and $G'/G''=H_1(\overline{S})$, where $\overline{S}$ is the *universal abelian cover* of $S$. That is, it is the cover associated to the abelianization homomorphism $G\to H_1(G)$, namely the cover with $\pi_1(\overline{S})=G'$ by the correspondence between subgroups and covers. A way to imagine the action of $G/G'$ on $G'/G''$ is that $H_1(S)$ acts on $H_1(\overline{S})$ as the group of deck transformations. We will characterize $H_1(\overline{S})$ as a $\mathbb{Z}[H_1(S)]$-module.

If $S$ is a compact surface with nonempty boundary, its fundamental group is the free group $F_k$ on some number of generators $k$. If $F_k=\langle g_1,g_2,\dots,g_k\rangle$ is the free generating set, the abelianization is a map $F_k\to \mathbb{Z}^k$ sending $g_i$ to a generator $t_i$. The ring $\mathbb{Z}[\mathbb{Z}^k]$ is isomorphic to $\mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$, the Laurent polynomial ring in $k$ variables with integer coefficients. I can't really explain the following calculations in enough detail here, other than what I am doing is calculating with $C_i(\overline{X_G})\cong \mathbb{Z}[H_1(G)]\otimes C_i(X_G)$ for a presentation complex $X_G$ of $G=\pi_1(S)$. For example, $t_ig_1$ represents the copy of $g_i$ (a distinguished lift of the $1$-cell for $g_i$ to $\overline{X_G}$) shifted by the deck transformation for $t_i$. (I might post a link to some notes about this at some point.) The complex only has $0$- and $1$-cells, so we need only identify the kernel of $\partial_1$, where $\partial_1(g_i)=t_i-1$. Since these are all pairwise coprime polynomials, $\ker\partial_1$ is minimally generated by $(t_j-1)g_i-(t_i-1)g_j$ for all $1\leq i<j\leq k$. Then
$$H_1(\overline{S}) \cong \bigoplus_{i=1}^{\binom{k}{2}} \mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$$
as a free $\mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$-module. As a special case, $k=1$ is when $S$ is an annulus, and we can see the above matches the observation that $H_1(\overline{S})=0$ since $\pi_1(S)$ is abelian.

In a more topological approach, we can replace $S$ with a wedge of $k$ circles $X=\bigvee_{i=1}^k S^1$. A model for $\overline{X}$ is to take $\mathbb{Z}^k\subset \mathbb{R}^k$ as the $0$-cells and take the length-$1$ line segments between these points as the $1$-cells. The kernel elements identified above are squares spanned by pairs of standard basis vectors.

If $S$ is a closed compact surface of genus $g$, then instead we have
$$G=\langle e_1,f_1,\dots,e_g,f_g \mid [e_1,f_1]\cdots[e_g,f_g]=1\rangle.$$
The abelianization is $\pi_1(G)\to \mathbb{Z}^{2g}$ with $e_i\mapsto s_i$ and $f_i\mapsto t_i$ for generators $s_1,t_1,\dots,s_g,t_g$.
\begin{align}
\partial_1 e_i &= s_i - 1\\
\partial_1 f_i &= t_i - 1\\
\partial_2([e_1,f_1]\cdots[e_g,f_g]) &= (1-t_1)e_1-(1-s_1)f_1 + \dots + (1-t_g)e_g-(1-s_g)f_g
\end{align}
The situation is very similar to the case of a surface with boundary in that the kernel of $\partial_1$ is the same, but now there is a single relation. This relation lets us write the element $(t_1-1)e_1-(s_1-1)f_1$ in terms of other elements of the generating set. Hence
$$H_1(\overline{S}) \cong \bigoplus_{i=1}^{\binom{2g}{2}-1} \mathbb{Z}[s_1^{\pm 1},t_1^{\pm 1},\dots,s_g^{\pm 1},t_g^{\pm 1}]$$
as a free $\mathbb{Z}[s_1^{\pm 1},t_1^{\pm 1},\dots,s_g^{\pm 1},t_g^{\pm 1}]$-module.

Therefore, the orbits of $G=\pi_1(S)$ on $G'/G''$ are in one-to-one correspondence with $\left(\binom{2g}{2}-1\right)$-tuples of Laurent polynomials in $2g$ variables with integer coefficients, modulo multiplication by $s_i$ and $t_i$ for all $1\leq i\leq g$.