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
People have pointed out that my hand-waving does not necessarily correspond to a proof. It was an absolute delight to revisit this proof. I have a new proof below, and it does not rely on the hand-wavy "minimality" hypothesis at all.
Let $x_i\in M_i$, and suppose $x_i\in D$. As you have noted, $x_i$ is a finite $A$-linear combination of elements of $C$ of the form $x_a-\mu_{ab}(x_a)$. Absorbing the coefficients from $A$ in the terms $x_a-\mu_{ab}(x_a)$, we get terms of the same form. So, let $i_1,i_2,\dots,i_k,j_1,j_2,\dots,j_k\in I$, $x_{(1)}\in M_{i_1}\dots,x_{(k)}\in M_{i_k}$ and suppose $i_r\le j_r$ for $r=1,2,\dots,k$ as well as $$x_i = (x_{(1)}-\mu_{i_1j_1}(x_{(1)}))+\dots+(x_{(k)}-\mu_{i_kj_k}(x_{(k)})).\qquad(1)$$ Since $\{i,j_1,j_2,\dots,j_k\}$ is a finite subset of $I$, which is directed, there exists $j_{\ast}$ such that $i\le j_{\ast}$ and $j_r\le j_{\ast}$ for $r=1,2,\dots,k$. Also, for $r=1,2,\dots,k$, we have $i_r\le j_r$, so $i_r\le j_{\ast}$.
For $a\in I$, let $\pi_{a}:C\to M_{a}$ be the homomorphism given by restricting the canonical projection $\prod_{b\in I}M_b\to M_{a}$ to $C$. On the one hand, we have $$\pi_a(x_i) = \begin{cases} x_i&\text{if $a=i$} \\ 0 &\text{if $a\ne i$}\end{cases}$$ Based on (1), we also find that $$\pi_a(x_i) = \sum_{i_b=a}x_{(b)} - \sum_{j_c=a}\mu_{i_cj_c}(x_{(c)}).$$ Then, $$\mu_{aj_{\ast}}(\pi_a(x_i)) = \sum_{i_b=a}\mu_{i_bj_{\ast}}(x_{(b)})-\sum_{j_c=a}\mu_{j_cj_{\ast}}(\mu_{i_cj_c}(x_{(c)}))$$ Summing over $a$, we have $$\sum_{a\in I}\mu_{aj_{\ast}}(\pi_a(x_i)) = \mu_{i_1j_{\ast}}(x_{(1)})-\mu_{j_1j_{\ast}}(\mu_{i_1j_1}(x_{(1)}))+\dots+\mu_{i_kj_{\ast}}(x_{(k)})-\mu_{j_kj_{\ast}}(\mu_{i_kj_k}(x_{(k)})).$$ The left side is $\mu_{ij_{\ast}}(\pi_i(x_i))$, and the right side is $0$. Therefore, $$\mu_{ij_{\ast}}(x_i) = 0.$$