Let $M$ be an $R$-module, and let $\{ M_i \}_{i \in I}$ be a family of $R$-modules. I need to prove that
$M \cong \bigoplus_{i \in I} M_i$ if and only if there exists families $\{ \mu_i \}_{i \in I}, \ \{ \rho_i \}_{i \in I}$ where $\mu_i \in Hom_{R-Mod}(M_i, M)$, $\rho_i \in Hom_{R-Mod}(M, M_i)$ such that
$1) \ \forall i \in I \ \ \rho_i \circ \mu_i = id_{M_i}$
$2) \forall i, j \in I, \ i \neq J : \ \ \rho_i \circ \mu_i = 0$
$3) \forall m \in M \ \ \rho_i(m) \neq 0$ for only finitely many $i \in I$
$4) \sum\nolimits_{i \in I} \mu_i \circ \rho_i = id_M$
If $M \cong \bigoplus_{i \in I} M_i$, then $M$ also satisfies the universal property(say, along with a family $\{ \mu \}_{i \in I}$ of canonical injections). In particular, for any $k \in I$ let $\{k_i \}_{i \in I}$ be a family of homomorphisms $k_i \in Hom_{R-Mod}(M_i, M_k)$. Let $k_i = 1_{M_k}$ if $i = k$ and let $k_i = 0$ if $i \neq k$. Then the universal property gives us a homomorphism $\phi_k: M \to M_k$ such that $\forall i \in I \ \phi_k \circ \pi_i = k_i$. So, $1)$ and $2)$ are taken care of. But how to prove that $3)$ and $4)$ also follow?
Besides, I currently have no idea how to prove that if $1) - 4)$ are satisfied then it's necessarily that $M$ is a coproduct of $\{M_i \}_{i \in I}$ in $R-Mod$.
Suppose first that indeed $M$ is isomorphic to $\bigoplus M_i$. There are projections and inclusions $\iota_i,\pi_i$ from and to each $M_i$ which satisfy conditions $1$ through $4$, and because such conditions are unaltered by conjugating your maps with the given isomorphism, you obtain one direction.
Conversely, assume such maps exist. Then there is a map $\rho : M\longrightarrow \bigoplus M_i$ such that $m\mapsto \sum \rho_i(m)$. Moreover, the maps $\mu_i$ assemble to give a map $\iota: \bigoplus M_i\longrightarrow M$. By construction and the given conditions, these are inverse isomorphisms.