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).