Let $G$ be a group and $M$ be a $G$-module, that is, an abelian group written additively on which $G$ acts: $$ (g,m)\mapsto g m.$$ We consider the group of coinvariants $$ M_G:=G/\langle g m -m\ |\ g\in G,\,m\in M\rangle. $$ Let $H\subseteq G$ be a subgroup of finite index. I am trying to understand the transfer map $ M_G\to M_H$.
Choose a section $s\colon H\backslash G\to G$ of the projection $G\to H\backslash G$ onto the quotient space $H\backslash G$. Consider the homomorphism $$ N\colon M\to M,\quad\ m\mapsto \sum_{x\in H\backslash G} s(x) m.$$ This homomorphism $N$ induces a homomorphism $$ N_*\colon M\to M_H, $$ which is clearly independent of the choice of the section $s$.
Question. Why does the homomorphism $N_*$ descend to a homomorphism $$ N_{**}\colon M_G\to M_H\ ?$$
In other words, for $m\in M$ and $g\in G$, why is $$ \sum_{x\in H\backslash G} s(x)\cdot( g m -m)$$ a linear combination of elements of the form $h m'-m'$ for $h\in H$ and $m'\in M$?
I will answer the concrete version of your question:
In this, I shall assume that $H\backslash G$ is the set of right cosets of $H$ in $G$. That is, an element of $H\backslash G$ has the form $Hg$ for some $g \in G$.
Now to your question: Let $m\in M$ and $g\in G$. Let $I$ denote the $\mathbb Z$-linear span of the elements of the form $h m'-m'$ for $h\in H$ and $m'\in M$. We must show that $$ \sum_{x\in H\backslash G} s\left(x\right)\cdot\left( g m -m\right) \in I .$$
To this purpose, I consider the map \begin{align} \phi : H\backslash G &\to H\backslash G, \\ x &\mapsto xg \qquad \left(\text{that is, } Hy \mapsto Hyg\right), \end{align} which is obviously bijective. For each $x \in H \setminus G$, we have \begin{align} s\left(x\right) gm - s\left(\phi\left(x\right)\right) m \in I . \label{eq.darij1.1} \tag{1} \end{align} [Proof: Let $x \in H \setminus G$. Set $y := s\left(x\right)$. Then, $x = Hy$, and thus $\phi\left(x\right) = \phi\left(Hy\right) = Hyg$ by the definition of $\phi$. Hence, $s\left(\phi\left(x\right)\right) = s\left(Hyg\right) \in Hyg$, so that $s\left(\phi\left(x\right)\right) = hyg$ for some $h \in H$. Considering this $h$, we now find \begin{align} \underbrace{s\left(x\right)}_{=y} gm - \underbrace{s\left(\phi\left(x\right)\right)}_{=hyg} m = ygm - hygm \in I \end{align} (since $h\in H$ and $ygm \in M$). This proves \eqref{eq.darij1.1}.]
Now, we can substitute $\phi\left(x\right)$ for $x$ in the sum $\sum_{x\in H\backslash G} s\left(x\right)\cdot m$ (since $\phi$ is a bijection). We thus obtain \begin{align*} \sum_{x\in H\backslash G} s\left(x\right)\cdot m = \sum_{x\in H\backslash G} s\left(\phi\left(x\right)\right)\cdot m . \end{align*} Hence, \begin{align*} & \sum_{x\in H\backslash G} s\left(x\right)\cdot\left( g m -m\right) \\ &= \sum_{x\in H\backslash G} s\left(x\right)\cdot gm - \underbrace{\sum_{x\in H\backslash G} s\left(x\right)\cdot m}_{=\sum_{x\in H\backslash G} s\left(\phi\left(x\right)\right)\cdot m} \\ &= \sum_{x\in H\backslash G} s\left(x\right)\cdot gm - \sum_{x\in H\backslash G} s\left(\phi\left(x\right)\right)\cdot m \\ &= \sum_{x\in H\backslash G} \underbrace{\left(s\left(x\right) gm - s\left(\phi\left(x\right)\right) m\right)}_{\in I \ \text{ (by \eqref{eq.darij1.1})}} \\ &\in I \qquad \left(\text{since $I$ is a $\mathbb{Z}$-module}\right). \end{align*} Qed.