A module $M$ is an internal direct sum of submodules $N_{1},\dots,N_{k}$ iff for all $m \in M$, there exists unique $n_{1} \in N_{1}, \dots, n_{k} \in N_{k}$ such that $m = n_{1}+n_{2}+\dots+n_{k}$
Now given a module $M$, and modules $N_{1}, N_{2}, \dots,N_{k} \subset M$, if $M$ isomorphic to the external direct sum of $N_{1}, \dots,N_{k}$, is $M$ the internal direct sum of $N_{1},\dots,N_{k}$?
I think if the isomorphism is $f(n_{1},\dots ,n_{k}) = n_{1} + \dots + n_{k}$ this is easy to prove. However, what if the isomorphism is defined differently?
No, this is not true in general. Consider $\mathbb Z$-modules $M=\mathbb Z\oplus \mathbb Z$, $N_1=2\mathbb Z\oplus \{0\}$, and $N_2=\{0\}\oplus 2\mathbb Z$.
Then $(a,b)\mapsto ((2a,0), (0,2b))$ gives an isomorphism $M\cong N_1\oplus N_2$, but $N_1+N_2=2\mathbb Z\oplus 2\mathbb Z\neq M$.