Consider two rings $R_1$ and $R_2$ each with their own identities, and their direct product $R_1 \times R_2$. It is well-known that the ideals of $R_1 \times R_2$ are all of the form $I_1 \times I_2$, where $I_1$ and $I_2$ are ideals of $R_1$ and $R_2$ respectively.
Now, let $R$ be a ring, and $M_1$ and $M_2$ be two modules over $R$. The question arises: Are all submodules of $M_1 \oplus M_2$, up to isomorphism, of the form $M' \oplus M''$ where $M'$ and $M''$ are submodules of $M_1$ and $M_2$ respectively?
To illustrate, let $k$ be a field. Consider the submodule $M := \{(x,x) \in k \oplus k \,|\, x \in k\}$ of $k \oplus k$. This submodule is not of the form $M' \oplus M''$ where $M'$ and $M''$ are submodules of $k$. However, $M$ is isomorphic to $k \oplus 0$ as module over $k$, and $k \oplus 0$ is a submodule of $k \oplus k$ over $k$. Therefore, this example does not serve as a counterexample to the posed question when considering isomorphism.
Here is a counterexample.
Let $R=k[x.y]/(x^2,y^2)$, where $k$ is a field. So $R$ is a four-dimensional algebra with basis $1,x,y,xy$.
Let $N=R/(xy)$. So $N$ is a three-dimensional $R$-module.
Then the map $N\to R\oplus R$ given by $n\mapsto(nx,ny)$ is injective, so $R\oplus R$ has a submodule isomorphic to $N$.
However $N$ is indecomposable (as it has a unique simple quotient), and is not isomorphic to a submodule of $R$ (as it has more than one simple submodule, but $R$ has only one).