Here's how I would present the construction given in Wikipedia.
Suppose $(M_i,\mu_{ij})$ is a directed system of modules. We begin by taking a disjoint union of the underlying sets of the $M_i$; in order to "keep them disjoint", the usual method is to "paint" each set with its index $i$ to ensure that if $i\neq j$, then the sets are disjoint. That is, we consider the set
$$\mathcal{M} = \bigcup_{i\in I}(M_i\times\{i\}).$$
The elements of $\mathcal{M}$ are pairs of the form $(x,i)$, where $i\in I$ and $x\in M_i$.
Note that $\mathcal{M}$ is not a module, at least not one with any natural structure: the operations we have on hand (the ones for the different $M_i$) are not defined on all of $\mathcal{M}$, they are only defined on proper subsets of $\mathcal{M}$.
We now define an equivalence relation on $\mathcal{M}$ as follows: $(x,i)\sim(y,j)$ if and only if there exists $k\in I$, $i,j\leq k$ such that $\mu_{ik}(x) = \mu_{jk}(y)$ in $M_k$. It is not hard to verify that this is an equivalence relation.
Let $\mathbf{M}$ be the set $\mathcal{M}/\sim$. Denote the equivalence class of $(x,i)$ by $[x,i]$.
We now define a module structure on $\mathbf{M}$: we define a sum on classes by the rule
$$ [x,i] + [y,j] = [\mu_{ik}(x)+\mu_{jk}(y),k]$$
where $k$ is any element of $I$ such that $i,j\leq k$. One needs to prove that this is well defined and does not depend on the choice of the $k$. Suppose first that $k'$ is some other element with $i,j\leq k'$. Let $\ell$ be an index with $k,k'\leq \ell$; then
$$\begin{align*}
\mu_{i\ell}(x) + \mu_{j\ell}(y) &= \mu_{k\ell}(\mu_{ik}(x))+\mu_{k\ell}(\mu_{jk}(y))\\
&= \mu_{k\ell}(\mu_{ik}(x) + \mu_{jk}(y)).
\end{align*}$$
Therefore, $[\mu_{ik}(x)+\mu_{jk}(y),k] = [\mu_{i\ell}(x)+\mu_{j\ell}(y),\ell]$. By a symmetric argument, we also have
$$[\mu_{ik'}x) + \mu_{jk'}(y),k'] = [\mu_{i\ell}(x) + \mu_{j\ell}(y),\ell],$$
so the definition does not depend on the choice of $\ell$.
To show it does not depend on the representative either, suppose $[x,i]=[x',i']$ and $[y,j]=[y',j']$. There exists $m$, $i,i'\leq m$ with $\mu_{im}(x)=\mu_{i'm}(x')$, and there exists $n$, $j,j'\leq n$, with $\mu_{jn}(y)=\mu_{j'n}(y')$. Pick $k$ with $m,n\leq k$. Then
$$\begin{align*}
[x,i]+[y,j] &= [\mu_{ik}(x)+\mu_{jk}(y),k]\\
&= [\mu_{mk}(\mu_{im}(x)) + \mu_{nk}(\mu_{jn}(y)),k]\\
&= [\mu_{mk}(\mu_{i'm}(x')) + \mu_{nk}(\mu_{j'n}(y')),k]\\
&= [\mu_{i'k}(x') + \mu_{j'k}(y'),k]\\
&= [x',i'] + [y',j'],
\end{align*}$$
so the operation is well-defined.
It is now easy to verify that $+$ is associative and commutative, $[0,i]$ is an identity (for any $i$) and that $[-x,i]$ is an inverse for $[x,i]$, so this operation turns $\mathbf{M}$ into an abelian group.
We then define a scalar multiplication as follows: given $r\in R$ and $[x,i]\in\mathbf{M}$, we let $r[x,i] = [rx,i]$. Again, one needs to show that this is well-defined (easier than the proof above), and verify that it satisfies the relevant axioms (not hard) to show that this endows $\mathbf{M}$ with the structure of a left $R$-module.
Now note that the maps $\mu_i\colon M_i\to \mathbf{M}$ given by $\mu_i(x) = [x,i]$ is a module homomorphism. The module $\mathbf{M}$ together with the maps $\mu_i$ are a direct limit of the system.
(The same construction works for Groups, Rings, etc).
There is no "linear extension" of the equivalence relation. Rather, we define an operation on $\mathcal{M}/\sim$, since $\mathcal{M}$ (being a disjoint union of the underlying set of the original modules) is not a module itself: it does not even have a total operation defined on it, just a bunch of partial operations.
Best Answer
Yes. It is straight forward to check the group axioms; they are inherited from the $G_i$. This construction can be found in every book which treats direct limits (which is the wrong name for directed colimits) and of course Wikipedia.