Let be $\{M_k,k\in I\}$ a directed system of $R$-modules. For every $i\in I$ consider injections $u_i:M_i\to\bigoplus_{k\in I}M_k$ and $\bigoplus_{k\in I}M_k\supseteq N$ generated by $u_jf_{ij}(x)-u_i(x).$ $(f_{ij}:M_i\to M_j)$
Let be $h_k:M_k\to N,\forall k\in I$ with $h_jf_{ij}=h_i$ for $j\ge i.$ Then there exist a unique $g:\left(\bigoplus_{k\in I}M_k\right)/N\to N$ such that $gu_k=h_k.$
My question is how could I define the map $g.$ My attempt was to set $h_k(x)=g\bigl([(\ldots,x,\ldots)]\bigr)$ where $[(\ldots,x,\ldots)]$ is a class containg an element $(\ldots,x,\ldots)$ and $x$ is on the $k$-th position.
However, I am not able to check if $g$ has needed universal property...
$g$ doesn't have a universal property. $g$ is the arrow the universal property the colimit of the directed system, which is $(\bigoplus_{k\in I} M_k)/N$, guarantees exists. However, you haven't fully specified that colimit and the condition $g\circ u_k = h_k$ doesn't make sense since the codomain of $u_k$, namely $\bigoplus_{k\in I}M_k$, doesn't match the domain of $g$, namely $(\bigoplus_{k\in I}M_k)/N$.
You have a quotient map $q:\bigoplus_{k\in I}M_k\to(\bigoplus_{k\in I}M_k)/N$ with which you can then define the colimiting cocone as the arrows $w_k:M_k\to(\bigoplus_{k\in I}M_k)/N$ defined by $w_k=q\circ u_k$. With this, the condition on $g$ is that $g\circ w_k = h_k$. However, even $h_k$ seems a bit suspect to me. We have for any $R$-module $T$ and collection of arrows $\{v_k:M_k\to T\mid k\in I\}$ such that $v_j \circ f_{ij} = v_i$, a unique arrow $p : (\bigoplus_{k\in I}M_k)/N\to T$ such that $p\circ w_k = v_k$. These are exactly the same constraints as are on $h_k$; the only difference is an arbitrary codomain is allowed. We can always choose $T=N$, so there's nothing wrong with $h_k$, but it is somewhat misleading and is not the full universal property of the colimit of the directed system. I will consider this more general case from this point on.
Abstractly, since we've given a concrete description of the colimit of the directed system, we can see the arrow we want induced by the universal properties of the pieces that make up the definition. In particular, given the family $\{v_k:M_k \to T\mid k \in I\}$, the universal property of the coproduct, $\bigoplus_{k\in I}M_k$, gives us a unique arrow $p' : \bigoplus_{k\in I}M_k\to T$ such that $p'\circ u_k = v_k$. Then, since $v_j\circ f_{ij} - v_i = 0$, we have, $$p'(u_j(f_{ij}(x))-u_i(x)) = p'(u_j(f_{ij}(x))-p'(u_i(x))= v_j(f_{ij}(x))-v_i(x) = 0$$ which means $p'$ satisfies the conditions needed to apply the universal property of the quotient by $N$. This gives us a unique map $p : (\bigoplus_{k\in I}M_k)/N\to T$ such that $p\circ q = p'$. Of course, combining this with the characterization of $p'$ gives us, $p\circ q\circ u_k = p'\circ u_k = v_k$.
We can get a concrete description of $p$ simply by concretely describing the arrows induced by the universal properties of the coproduct and quotient, $p'$ and $p$. To do this, we need a concrete description of the $R$-modules $\bigoplus_{k\in I}M_k$ and $(\bigoplus_{k\in I}M_k)/N$. I will describe $\bigoplus_{k\in I}M_k$ as formal (finite) sums of pairs $(i,x)$ where $i\in I$ and $x\in M_i$, with the operations satisfying $r(i,x)=(i,rx)$, $(i,0) = 0$, and $(i,x)+(i,y)= (i,x+y)$ in addition to the other properties required of the operations of an $R$-module. For $i\neq j$, $(i,x)+(j,y)$ just doesn't "simplify", just like with complex numbers $1+i$ doesn't simplify further. You should show that this description is isomorphic to whichever concrete description you like to use. We can then define $u_k:M_k \to \bigoplus_{k\in I}M_k$ as $u_k(x)=(k,x)$ which can easily be verified to be an $R$-module homomorphism. $p'$ is then simplify defined as $p'((i,x))= v_i(x)$ and extended to all formal sums by linearity. We clearly have $p'\circ u_k = v_k$.
As is typical for quotients, we can concretely represent $(\bigoplus_{k\in I}M_k)/N$ as being $\{\{m+n\mid n\in N\}\mid m\in\bigoplus_{k\in I}M_k\}$. As usual, we can write the equivalence class of $m$, as $m+N = \{m+n\mid n\in N\}$. The arrow $q$ is then $q((i,x))=(i,x)+N$ extended to all formal sums by linearity. $p$ is then defined by $p(m+N)=p'(m)$ which clearly gives $p \circ q = p'$, but we have to check that this definition is well-defined meaning given any other $m'\in m+N$, $p'(m')=p'(m)$. By definition, $m'=m+n$ for some $n\in N$, and by definition of $N$ this means $n=\sum_{(i,j)\in S}(n_j,f_{n_in_j}(x_{(i,j)}))-(n_i,x_{(i,j)})$ where $S\subseteq I\times I$ is finite. So the well-definedness condition is $$p'(m+\sum_{(i,j)\in S}(n_j,f_{n_in_j}(x_{(i,j)}))-(n_i,x_{(i,j)}))=p'(m)$$ or $$\sum_{(i,j)\in S}\left[p'((n_j,f_{n_in_j}(x_{(i,j)})))-p'((n_i,x_{(i,j)}))\right]=0$$ which is, by definition of $p'$, $$\sum_{(i,j)\in S}\left[v_{n_j}(f_{n_in_j}(x_{(i,j)}))-v_{n_i}(x_{(i,j)})\right]=0$$ which holds by assumption.
Bringing everything together, elements of $(\bigoplus_{k\in I}M_k)/N$ look like $\sum_{j\in J}(j,x_j)+N$ for finite subsets $J\subseteq I$ and $p(\sum_{j\in J}(j,x_j)+N) = \sum_{j\in J}v_j(x_j)$.
As some minor nits with wording, saying "consider [the] [...] $\bigoplus_{k\in I}M_k\supseteq N$" is unnatural. It's like saying "consider the three greater than or equal to $x$". It would make more sense to flip it, e.g. "consider the $N \subseteq\bigoplus_{k\in I}M_k$" i.e. "consider the $x$ less than or equal to three". Similarly, "[given $h$] set $h_k(x)=g([(\dots,x,\dots)])$" also reads poorly. Again, an analogy would be "given a number $x$, set one to be $x$", and again the solution is just to flip them around, i.e. $g([(\dots,x,\dots)])=h_k(x)$