Previously I asked a questionon whether isomorphic to external direct sum implies "is internal direct sum". The answer turns out to be no, but whenever the internal direct sum is "defined", it is isomorphic to the internal direct sum. Sometimes the internal direct sum is not always defined, but I cannot think of an example. More precisely:
Given module $M$ and submodule $N_{1},\dots ,N_{k}$ suppose $M$ isomorphic to the external direct sum of $N_{1},\dots ,N_{k}$, can there not exists a submodule $N \subset M$, such that $N$ is the internal direct sum of $N_{1},\dots,N_{k}$?
Of course, if $N_{i} = N_{j} \neq 0$ for some $i,j$, there is no way for the internal direct sum make sense. However, I cannot conceive a concrete example that satisfies the condition $M \cong N_{1} \oplus \dots \oplus N_{k}$.
The key is to take a module that has a repeated direct summand, but choose $N_1$ and $N_2$ so they both "take" the same copy of that summand. That way the intersection $N_1\cap N_2$ is nontrivial, so there can be no "internal direct sum" of $N_1$ and $N_2$.
The simplest example is to take $M=R\oplus R$ and take $N_1=N_2=R\oplus\{0\}$. Since $N_i\cong R$, we get that $M$ is isomorphic to the external direct sum $N_1\oplus N_2$, but not to an internal direct sum because $N_1\cap N_2=R\oplus\{0\}\neq \{(0,0)\}$.
Alternatively, if you want $N_1\neq N_2$, then take $M=R\oplus R\oplus R\oplus R$, and let $N_1=R\oplus R\oplus\{0\}\oplus \{0\}$, and $N_2=R\oplus \{0\}\oplus \{0\}\oplus R$.