Direct sum of modules Let $R$ be a PID,$M$ an $R$-module and $M$ is the direct sum of $M_1, \dots, M_k$ where $M_i \leq M$ for $1 \leq i \leq k$.
Now let $N$ be a submodule of $M$ .
Is it true that $N$ is the direct sum of $N_1,\dots,N_t$ where $N_i \lt N$ and $N_{k-t+1} \le M_i$?
 A: Take $R=\mathbb{Z}$, $M=\mathbb{Z} \oplus \mathbb{Z}= M_1 \oplus M_2$ and $N=\mathbb{Z}\cdot (1,1)$.
$N\cap M_1 =N\cap M_2 = 0$ so it can't be a direct sum of submodules of $M_1, M_2$
A: If I understand the question correctly then
$$
\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C
$$
is a counterexample (here if have chosen $R=\mathbb C$).
A: For a general module M over a PID R, its submodules need not even be isomorphic to direct sums of submodules of its proper direct summands.  For instance, the Z-module Q10 is a direct sum of proper submodules, but it has directly indecomposable submodules of rank 10 (so they cannot be written as any non-trivial direct sum).  Since the torsion-free rank of a submodule can only decrease, this means M = Q⊕Q9 is a counterexample.
If we try to do what the question asks, and get a direct sum decomposition from submodules of the direct summands, then things go wrong even for M=Z/2Z ⊕ Z/2Z and the submodule N={(x,x):x in Z/2Z }.  N is directly indecomposable, and is not a submodule of the specified direct summands of M (though it is itself a summand in a different decomposition).
