Let $M$ be a module over a ring $R$, we define $Gen(M)$ as the class of all modules over $R$ which are generated by $M$, that means that if $N \in Gen(M)$ there is an epimorphism $f:M^{(X)} \twoheadrightarrow N$ where $M^{(X)}=M \oplus M \oplus M \oplus...$ ,$X$ times and $X$ is an arbitrary set. I want to prove that $ \bigoplus_{i \in I} N_{i} \in Gen(M)$ if $N_{i} \in Gen(M)$ for each $i \in I$, for $I$ arbitrary set.
I was thinking in an easier case just to gain more intuition. So lets suppose we have $N_{1}, N_{2} \in Gen(M)$ that means we have two epimorphisms without lost of generality $f:M \oplus M \twoheadrightarrow N_{1}$ and $g:M \twoheadrightarrow N_{2}$, the only idea I got to construct an epimorphism from direct copies of $M$ into $N_{1} \oplus N_{2}$ was $\phi:M \oplus M \oplus M \twoheadrightarrow N_{1} \oplus N_{2}$ as $\phi(m_{1},m_{2},m_{2}):=(f(m_{1},m_{2}),g(m_{3})) \in N_{1} \oplus N_{2}$. Almost sure this morphism is linear,well defined and surjective. I was thinking in generalize this idea to create an epimorphism from $M^{(X)}$ to $\bigoplus_{i \in I} N_{1}$ where each $N_{i} \in Gen(M)$ and $I$ is an arbitrary set.
Your approach is correct. For if $$M^{(X_i)}\to N_i\to\{0\}$$ is an exact sequence for every $i\in I$, then $$\bigoplus_{i\in I}M^{(X_i)}\to\bigoplus_{i\in I}N_i\to\{0\}$$ is exact as well. Since $\bigoplus_iM^{(X_i)}\cong M^{(X)}$ where $X=\coprod_iX_i$ (disjoint union), we get $\bigoplus_iN_i\in\operatorname{Gen}(M)$.