I just started studying modules and stumbled upon the fact that:
the direct sum of a family of modules is the submodule generated by their union: $$\bigoplus_{i\in I}A_{i}=\langle \bigcup_{i\in I}A_{i} \rangle$$
I could prove this statement for a finite or countable index set $I$, but I don't know how a proof would look for an uncountable set.
Also, do those modules need to satisfy the following propety in order for the proposition to be valid? $$M_{j}\cap \langle \underset{i\ne j}{\bigcup_{i\in I}} M_{i}\rangle=0$$
Note that an element of $\bigoplus_{i \in I} A_i$ is a finite sum of elements of the $A_i$, i.e., almost all terms $a_i$ of $(a_i)_{i \in I} \in \bigoplus_{i \in I} A_i$ are zero. Hence such an element is easily in $\left\langle \bigcup_{i \in I} A_i\right\rangle$: We can write $$(a_i)_{i \in I} = \sum_{i \in I, a_i \neq 0} (\underbrace{\delta_{ij}a_i}_{= \begin{cases}0 & j \neq i \\ a_i & j=i \end{cases}})_{j \in I}$$ and the summand at index $i \in I$ lies in $A_i$ (read: the image of $A_i$ in $\bigoplus_{i \in I} A_i$). Hence $(a_i)_{i \in I}$ is generated by all $A_i$ together (again read: their imagas in $\bigoplus_{i \in I} A_i$).
A possible source of confusion is that, as I said, the direct sum only contains elements where the non-zero part is finite (I guess you aren't aware of this because you said something about proofs for finite/countable sets.)
As you can see my proof didn't use any further assumptions, so the answer to your second question is no. The confusion here probably stems from notation: With $A_i$ we actually mean the image of the natural homomorphism $A_i \to \bigoplus_{i \in I} A_i$. Then $A_i$ is a submodule of $\bigoplus_{i \in I} A_i$. The task then is just that all those submodules together generate the whole module $\bigoplus_{i \in I} A_i$.
It is not something like that: Given some module $M$ and submodules $A_i$, show that $M = \bigoplus A_i$. For such a statement you would need assumptions on the $A_i$, similar to what you suggested.