Let $N \subseteq M$ be modules over a Noetherian integral domain $R$. Assume that if $m \in M$ and $rm \in N$, then $rm = 0$. Let $P \subseteq F$ be free $R$-modules of finite rank, and suppose $f: M \rightarrow F$ and $g: N \rightarrow P$ are surjective $R$-module homomorphisms such that
$$\begin{array} M M& \rightarrow & F \\ \cup & & \cup \\ N & \rightarrow & P \end{array}$$
is commutative. Let $s_1: P \rightarrow N$ and $s_2: F \rightarrow M$ be $R$-module homomorphisms such that $g \circ s_1 = 1_F$ and $f \circ s_2 = 1_P$. Is it possible to choose $s_1$ and $s_2$ so that $s_1 = s_2|_P$? What if the quotient $F/P$ is assumed to be projective?
Here is the motivating example I have in mind for this question: $M \supseteq N$ are locally compact abelian topological groups, with respective maximal compact subgroups $O_M$ and $O_N = O_M \cap N$. The quotients $M/O_M$ and $N/O_N$ are finite rank discrete free abelian groups, with $N/O_N \subseteq M/O_M$ naturally. I would like to rewrite $M$ and $N$ as direct products of topological groups
$$M = O_M \times A \space \space \space \space \space \space \space \space N = O_N \times B$$
for finite rank free abelian groups $A \supseteq B$, such that $N$ sits inside $M$ as $B$ sits inside $A$.
[I am going to ignore the assumption stated in the second sentence of the question, since as I commented it makes the question trivial, and the question has an interesting answer without it.]
This is not true in general. For instance, let $R=\mathbb{Z}$, $F=\mathbb{Z}$, $P=2\mathbb{Z}$, $M=\mathbb{Z}^2$, and $N=\{(a,0):a\in\mathbb{Z}\}\subset M$. Let $f(a,b)=2a+b$, so its restriction $g(a,0)=2a$ maps $N$ onto $P$. Then there is only one choice of $s_1$, namely $s_1(c)=(c/2,0)$, and this does not extend to a homomorphism $s_2:F\to M$.
On the other hand, it is true if $F/P$ is projective. In that case, the quotient map $F\to F/P$ splits, so $P$ is a direct summand of $F$; say $F=P\oplus Q$. We can then just take any $s_1$ and extend it to $s_2$ by defining $s_2$ to be a right inverse to $f$ on $Q$ (this is possible since $Q$ is projective).