Assume that $R$ is and integral domain.
If $M$ is an $R$-module that contains a submodule $N$ free of rank $n$, such that $M/N$ is a torsion $R$-module, then $M$ is of rank $n$.
My reasoning: That $M/N$ is a torsion-module implies that for each $y_i \in M \setminus N$ there exists a non-zero $r_i \in R$ s such that
$$r_iy_i = r^{i}_1x_1+ \ldots + r^{i}_nx_n$$ where $x_1,\ldots,x_n$ is a basis for $N$.
Now, let´s choose arbitrary $y_1,\ldots,y_{n+1} \in M \setminus N$.
We know that since $$N \cong R^n$$ then $$\{e_1,\ldots,e_n\}$$ is a basis for $N$.
We can then write $$r_iy_i = a_{1i}e_1 + \ldots + a_{ni} e_n$$
where $r_i$ is choosen as the annihilator to $\bar{y_i} \in M/N$.
Now,we get a matrix $A$ of dimension $$(n+1) \times (n+1)$$ such that the last row is constantly $0$, where the columns $A_i$ are made up of the coefficients for $r_iy_i$ in the basis $\{e_1,\ldots,e_n\}$.
By computing the determinant from the last row:s elements as coefficients for the minors, we see that $$\operatorname{det}(A) = 0$$ and it follows from lemma that the columns are $R$-linearly dependent.
I.e. there exists atleast two columns $A_i,A_j$ for $i \neq j$ such that $rA_i = A_j$ for some $r \in R$ (I think).
Since $$rA_i = r(a_{1i}e_i+ \ldots + a_{ni}e_n) = r(r_iy_i) = a_{1j} e_1+ \ldots + a_{nj} e_n = A_j = r_jy_j$$ it follows that $$(rr_i)y_i-r_jy_j= = 0$$ but I think by assumption $r_i$ and $r_j$ are non-zero, and since $R$ is an integral domain, both $rr_i$ and $r_j$ are non-zero, so that we find that $y_i,y_j$ are $R$-linearly dependent.
Now, this extends to elements in $N$ ofc, since $N$ is free.
Is this argument correct? I feel I could possibly make it better; for example, by not dividing it into the case $M\setminus{N}$ and $N$, and also my argument for why $r_i$ and $r_j$ are non-zero.
One fact that is often helpful for reasoning about rank is that the rank of $M$ is thr dimension of $K \otimes_R M$ over $K$, where $K$ is the fraction field of $R$. Using this, one can reduce some questions about rank to linear algebra.
For example, if $0\to M' \to M \to M''\to 0$ is exact, then tensoring with $K$, which is flat over $R$, we obtain $$0\to K\otimes_R M' \to K \otimes_R M \to K \otimes_R M'' \to 0$$ Which implies $\operatorname{rank}(M)=\operatorname{rank}(M')+\operatorname{rank}(M'')$.
This implies the statement in question, even without the freeness assumption, because torsion modules have rank zero.