A left $R$-module is an abelian (and thus additive) group $V$ on which a ring $R$ acts. The behavior of these actions respects the ring structure, in other words we have linearity in both directions, unity as the identity map and associativity (which is to say the ring's multiplication is composition):
$$(r+s)\cdot v=r\cdot v+s\cdot v, \qquad r\cdot(x+y)=r\cdot x+r\cdot y,$$
$$1_R\cdot v=v, \qquad (rs)\cdot v=r\cdot(s\cdot v), \quad \forall r,s\in R,~~v,x,y\in V.$$
Since each action amounts to an endomorphism of $V$, we may identify $R$-actions with a subring of the ring of endomorphisms, $\mathrm{End}(V)$. Thus, just as $\rho:G\to\mathrm{GL}(V)$, we have $\rho:R\to\mathrm{End}(V)$ to help us "represent" how $R$ acts on $V$. (Notice a vector space $V$ is an example of the additive group.)
This is similar to the usual notion representation, except now we can add actions together instead of simply multiply (compose) them. We are admitting more structure, endomorphisms instead of just automorphisms. There is an intuitive sense in which a vector space over $k$ with representation of $G$ is essentially the same as $V$ interpreted as a $kG$-module: given one, we can easily go to the other.
Given $V$ and $G$, we can extend the actions of $G$ to actions of $kG$ by admitting linearity:
$$\left(\sum_{g\in G} a_g g\right) v:=\sum_{g\in G}a_g(g\cdot v) \qquad \forall v\in V, ~a_g\in k.$$
Conversely, $G$ is a (not necessarily proper) subgroup of $(kG)^\times$ (the group of units, or invertible elements, in the group algebra $kG$), so given $V$ as a $kG$-module, we can make it a vector space over the field $k$ by defining scalar multiplication via $av=(ae)\cdot v$, where $a\in k$ and $e\in G$ is identity, and then it is automatically a representation of $G$ by letting $g\in G$ act via $g\cdot v$ as a $kG$-action.