Problem: Let $\{N_i\}_{i \in I}$ be a family of $R$-submodules of $M$. The following are equivalent.
- The map $\pi: \bigoplus_{i \in I} N_i \rightarrow \langle \cup_{i \in I} N_i \rangle$ defined by $(m_i)_{i \in I} \mapsto \sum_{i \in I} m_i$ is an $R$-module isomorphism.
- For any finite subset $\{i_1,i_2,\dots,i_k\}$ of $I$ one has $N_{i_1} \cap (N_{i_2} + \dots +N_{i_k}) = \{0\}$.
- For every element $x \in \langle \cup_{i \in I} N_i \rangle$ there are unique elements $a_i \in N_i$ for all $i \in I$, finitely many non-zero, such that $x = \sum_{i \in I}a_i$.
The finite case is straightforward, but I'm not sure if the proofs extend directly to the arbitrary case. In particular I imagine the $(2 \implies 3)$ argument may have issue, since I wasn't sure I was indexing properly. The issue I had was making precise the idea that any element of $\langle \cup_{i \in I} N_i \rangle$ is already a finite sum $x = \sum_1^n a_i$ where $a_i \in \cup_{i \in I} N_i$ and so it's obvious that it corresponds to some sequence in the direct sum where (informally) you've filled in missing coordinates with zero, and collected elements in the same submodules together. Let me know how I might be able to polish this argument or fix any errors I've made.
Attempt:
$(1 \implies 2)$. Suppose for contradiction there is a finite subset $\{i_1,\dots,i_k\} \subseteq I$ for which $N_{i_1} \cap (N_{i_2}+\dots+N_{i_k}) \neq \{0\}$. There exists some non-zero $a_{i_1} \in N_{i_1}$ which can be written as $a_{i_1} = a_{i_2}+\dots+a_{i_k}$ with $a_{i_j} \in N_{i_j}$ for all $j$. It follows that $a_{i_1}-a_{i_2}-\dots-a_{i_k} = 0$ and therefore corresponds to a non-zero element of the direct sum which is mapped to $0$, and hence $\ker\pi \neq 0$.
$(2 \implies 3).$ By definition $x \in \langle \cup_{i \in I} N_i \rangle$ means that $x = \sum_1^n a_{i_k}$ where $a_{i_k} \in \cup_{i \in I} N_i$ for all $1 \leq k \leq n$. In particular this means that we may write $x = \sum_{i \in I} a_i$ with $a_i \in N_i$ non-zero whenever $i \notin \{i_1,\dots,i_n\}$. Now if $x = \sum_1^n a_{i_k} = \sum_1^m b_{i_j}$ where $a_{i_k},b_{i_j} \in \cup_{i \in I} N_i$, assuming without loss of generality that $n>m$ we may write $\sum_1^n (a_{i_k}-b_{i_j}) = 0$ where $b_{i_j} = 0$ for all $j \geq m$. By assumption, for the collection $\{i_1,\dots, i_n\}$ one has that $N_{i_1} \cap (N_{i_2}+\dots+N_{i_n}) = 0$ therefore $a_{i_k}-b_{i_j} \in N_{i_k}$ for all $1 \leq k \leq n$ it follows that $a_{i_k}-b_{j_k} =0 \implies a_{i_k} = b_{i_j}$ for all $1 \leq k \leq n$. Therefore the representation of $x = \sum_{i \in I} a_i$ is unique.
$(3 \implies 1)$. The map $\pi$ is clearly a surjective module homomorphism since any finite sum $\sum_1^n a_i$ with $a_i \in N_i$ obviously corresponds to the image of some sequence $(a_i)_{i \in I} \in \oplus_{i \in I} N_i$ under $\pi$. The assumption that any $x \in \langle \cup_{i \in I} U_i \rangle$ is uniquely represented as $\sum_{i \in I} a_i$ clearly implies $\pi$ injects since if $\pi(a_i) = \pi(b_i)$ then $\sum_{i \in I} a_i = \sum_{i \in I} b_i \implies a_i = b_i$.
All your arguments are correct, but, to be honest, the proof of $(\mathbf2\!\implies\!\mathbf3)$ is a bit complicated to understand. You can avoid it by instead proving $(\mathbf1\!\iff\!\mathbf3)$ and $(\mathbf2\!\iff\!\mathbf3)$.
Lemma: $\left\langle \bigcup_{i \in I} N_i \right\rangle = \left\{ \sum_{i \in I} m_i : (m_i)_{i \in I} \in \bigoplus_{i \in I} N_i \right\}$.
Proof: For every $(m_i)_{i \in I} \in \bigoplus_{i \in I} N_i$, $\sum_{i \in I} m_i \in \left\langle \bigcup_{i \in I} N_i \right\rangle$; that is, the left hand side contains the right hand one. Also, it is easy to see that the right hand side is an $R$-submodule of $M$ that contains $\bigcup_{i \in I} N_i$; so, the right hand side contains the left hand one. $\square$
Therefore, by the above lemma, $\pi$ is a surjective $R$-module homomorphism; so the statement $\bf1$ is equivalent to simply saying that "$\pi$ is injective". In turn, it is easy to see that the latter (together with the lemma) is equivalent to $\bf3$.
Now, note that in $(\mathbf1\!\implies\!\mathbf2)$ you really proved that "$\pi$ is injective" implies $\bf2$. Let’s see that $\bf2$ implies that "$\pi$ is injective":
Given $(m_i)_{i \in I} \in \ker \pi$, let $i_1,i_2,\dots,i_k$ be all the elements of the finite set $\{i \in I : m_i \neq 0\}$. Since $$ 0=\pi \left( (m_i)_{i \in I} \right) = \sum_{i \in I} m_i = m_{i_1}+m_{i_2}+\cdots+m_{i_k} $$ we have that $$ -m_{i_1} = m_{i_2}+\cdots+m_{i_k}. $$ The above element belongs to $N_{i_1} \cap (N_{i_2}+\cdots+N_{i_k})$; so that, by $\bf2$, it is equal to $0$. Thus, $m_{i_1}=0$ and $m_{i_2}+\cdots+m_{i_k} = 0$.
By an inductive argument, it follows that all the coordinates of $(m_i)_{i \in I}$ are zero. Hence, $\pi$ is injective.