Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?

Nice question. No, this is not true in general, even if $R$ is a field.

Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_\infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.

Let $R$ be a nontrivial (necessarily noncommutative) ring with the property that $R\cong R\oplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism ring of an infinite-dimensional vector space.) Then $R$ has a proper submodule $N$ with $N\cong R$ and $R/N \cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$. Clearly $N_1\not\cong N_2$.