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.
Denote the direct limit of $\left\{G_\alpha , f_\alpha^\beta \right\}$ as $G$, and the direct limit of $\left\{H_i,g_i^j\right\}$ as $H$. Then we claim that $G\oplus H$ is the direct limit of $\left\{G_\alpha \oplus H_i, f_\alpha^\beta \oplus g_i^j \right\}$. And your questions are answered by this claim.
The morphisms $G_\alpha\oplus H_i\rightarrow G\oplus H$ are induced by the morphisms $G_\alpha\rightarrow G$ and $H_i\rightarrow H$.
Suppose there is a compatible system of morphisms $G_\alpha\oplus H_i\rightarrow M$. Since $\operatorname{Hom}(G_\alpha\oplus H_i,M)\cong\operatorname{Hom}(G_\alpha, M)\oplus\operatorname{Hom}(H_i, M)$, this induces compatible systems of morphisms $G_\alpha\rightarrow M$ and $H_i\rightarrow M$. Then by the universal property of the direct limit, these morphisms factor through $G_\alpha\rightarrow G$ and $H_i\rightarrow H$. This immediately implies that the morphisms $G_\alpha\oplus H_i\rightarrow M$ factor through $G_\alpha\oplus H_i\rightarrow G\oplus H$. Therefore $G\oplus H$ is the direct limit of $\left\{G_\alpha \oplus H_i, f_\alpha^\beta \oplus g_i^j \right\}$ as claimed.
Hope this helps.
Best Answer
If by "directed limit" you mean "filtered colimit" (which I imagine is the case here) then the answer will always be yes, the main reason being that filtered colimits commute with finite limits in $\mathbf{Ab}$. $\require{AMScd}\def\id{\operatorname{id}}$
If $A:J\to\mathbf{Ab}$ is a filtered diagram (i.e., directed system) and $B$ is a subdiagram (this is just my way of phrasing what you describe in your question), then like you mentioned the inclusion $i:B\Rightarrow A$ induces a morphism $i:\varinjlim B\to\varinjlim A$ of colimits.
Now, this is a monomorphism (i.e., injective) iff \begin{CD} \varinjlim B @>\id>> \varinjlim B \\ @V\id VV @VViV \\ \varinjlim B @>>i> \varinjlim A \end{CD} is a pullback square, which is true because this is the colimit of the system of pullback squares \begin{CD} B_j @>\id>> B_j \\ @V\id VV @VV\subseteq V \\ B_j @>>\subseteq> A_j \end{CD} for $j\in J$.
If "directed limit" just meant an arbitrary colimit, the answer is no: for example, $\Bbb Z$ is the colimit of the diagram $\Bbb Z\xleftarrow{\id}\Bbb Z\xrightarrow{\id}\Bbb Z$, but if we look at the subdiagram $\Bbb Z\leftarrow0\rightarrow\Bbb Z$ the colimit is $\Bbb Z\oplus\Bbb Z$, and the induced map $\Bbb Z\oplus\Bbb Z\to\Bbb Z$ is given by $(n,m)\mapsto n+m$, which is certainly not injective (and $\Bbb Z\oplus\Bbb Z$ simply cannot be a subgroup of $\Bbb Z$).
To see this more concretely, the direct limit $\varinjlim A$ as the result of putting all the elements of every $A_j$ together, identifying elements via the transition maps. If the diagram is not filtered, then it is possible that "the reason $a_j\in A_j$ and $a_{j'}\in A_{j'}$ sum to zero is that they both come from some $A_k$ with $k\leq j,j'$ and in there, they sum to zero" so if your subgroup $B_k$ does not include $a_j$ and $a_{j'}$, then these elements may not sum to zero in $\varinjlim B$.
However, with a filtered diagram, any $j,j'$ have an $\ell\geq j,j'$, meaning for any $a_j\in A_j$ and $a_{j'}\in A_{j'}$ that we can find these elements in $A_\ell$ and perform the addition there. Therefore, if $a_j+a_{j'}=0$, this fact will be witnessed in some group in the directed system containing both of those elements. This resolves the original issue because if $a_j\in B_j$ and $a_{j'}\in B_{j'}$ as well, then they must both lie in $B_\ell\subseteq A_\ell$ and therefore they will still sum to zero.
In other words, an element of $\varinjlim A$ is zero precisely when it is the image of the zero in one of the $A_{j'}\to\varinjlim A$. If $\phi:\varinjlim B\to\varinjlim A$ sends some $b$ to zero, note that $b$ comes from some $B_j$, and $\phi(b)$ is the image of some zero in $A_{j'}$. Take $\ell\geq j,j'$, then $b$ gets sent via the transition map $B_j\to B_\ell$ to the same element that $0\in A_{j'}$ is sent to via its transition map (which is again $0$). That is, $b$ vanishes in $B_\ell$, and therefore $b$ must have already been zero in $\varinjlim B$, showing that $\phi$ is injective.