Let $\{(M_i, N_i)\}_{i\in I}$ be a collection of $R$-modules where $N_i$ is a submodule of $M_i$ for each $i$ in a finite indexing set $I$. Then we can show
$$\bigoplus_{i\in I} \frac{M_i}{N_i} \simeq \frac{\bigoplus_{i\in I}M_i}{\bigoplus_{i\in I}N_i}$$
with the following argument.
Let $M, N$ be $R$-modules and $A$ and $B$ submodules of $M$ and $N$ respectively. Consider the map $f:{M\oplus N}\to{(M/A)\oplus(N/B)}$ given by $f(m, n) = (m + A, n + B)$. We see that $f$ is the direct sum of the canonical quotient maps $\pi_A:{M}\to{M/A}$ and $\pi_B:{N}\to{N/B}$, so $f$ is a well-defined surjective module homomorphism.
Suppose $(m, n)\in \mbox{ker}(f)$, then $f(m, n) = (\pi_A(m), \pi_B(n)) = (0, 0)$ and therefore $m\in A$ and $n\in B$. Thus $\text{ker}(f)\subset A\oplus B$. Now, suppose $(m, n)\in A\oplus B$, then clearly $f(m, n) = 0$ and therefore $\text{ker}(f) = A\oplus B$. Then applying the first isomorphism theorem we have
$$ \frac{M}{A}\oplus\frac{N}{B}\simeq \frac{M\oplus N}{A\oplus B}$$
Therefore, for the finite collection $\{(M_i, N_i)\}_{i\in I}$ we can recursively apply the identification above to obtain
$$\bigoplus_{i\in I} \frac{M_i}{N_i} \simeq \frac{\bigoplus_{i\in I}M_i}{\bigoplus_{i\in I}N_i}$$
What I would like help with is how can this argument be extended to an infinite sum
$$\bigoplus_{i=0}^\infty\frac{M_i}{N_i} \stackrel{?}{\simeq} \frac{\bigoplus_{i=0}^\infty M_i}{\bigoplus_{i=0}^\infty N_i}$$
if it can be extended at all? And how would the argument change/break down for an infinite product $\prod_{i =0}^\infty$ ? I've always struggled with reasoning behind infinite sums/infinite products arguments, so help understanding this would be greatly appreciated! Thank you.