In abelian categories we know finite products and coproducts are isomorphic, but they are not necessarily the same when it comes to infinite products and coproducts, so you can't really get a unique morphism from an object $Z$ to $\coprod_{i \in I} A_i$ given an arbitrary family $\{f_i\}_{i \in I}$ of morphisms $f_i$ from $Z$ to $A_i$ if $I$ is infinite, while you can do so when $I$ is finite (using the fact that it is a coproduct).
However, in the category of abelian groups if only finitely many of those morphisms $f_i$ are non-zero, then there is such a special morphism (namely, the one that takes $z \in Z$ and sends it to $\sum_{i \in I} f_i(z)$).
So I was wondering, is it true that, given a family $\{f_i\}_{i \in I}$ of morphisms $f_i$ from $Z$ to $A_i$, there is a unique morphism $f$ from $Z$ to $\coprod_{i \in I}$ such that $p_i \circ f = f_i$ for every $i \in I$, with $p_i$ being the projection from $\coprod_{i \in I} A_i$ to $A_i$ (more explicitly, the unique morphism such that $p_j \circ u_i = 1_{A_i}$ if $i = j$ and $p_j \circ u_i = 0$ if $i \neq j$ if $u_i$ is the inclusion of $A_i$ on $\coprod_{i \in I} A_i$) if, and only if, $f_i \neq 0$ for finitely many $i \in I$?
No. For example, $Z$ could itself be the infinite coproduct $\bigsqcup_{i \in I} A_i$, with the maps $f_i : Z \to A_i$ given by the projections $p_i$. Then there is a morphism $f : Z \to \bigsqcup_{i \in I} A_i$ such that $p_i \circ f = f_i$, namely the identity, and none of the $f_i$ are zero (assuming the $A_i$ are nonzero).
Even in the category of abelian groups you can see that in order for $\sum_i f_i(z)$ to make sense it's not necessary that only finitely many of the $f_i$ are nonzero, only that for any $z \in Z$ only finitely many of the $f_i(z)$ are nonzero, and that condition is satisfied by the infinite coproduct as above.
However, this is true if $Z$ is compact, meaning that $\text{Hom}(Z, -)$ preserves filtered colimits. Since the infinite coproduct is the filtered colimit of the finite coproducts, it follows that any map $Z \to \bigsqcup_{i \in I} A_i$ factors through a finite coproduct.
In $\text{Mod}(R)$ the compact objects are exactly the finitely presented modules. In a general abelian category they are the objects which "behave as if they are finitely presented."