I'm wondering why it's important that every module is a quotient of a free module.
In every text I've referred to, it seems to be an important theorem, but I'm failing to see how it's actually useful. If someone could provide some motivation behind this theorem, that would be very helpful.
To begin with, note that an isomorphism $M\simeq L/N$, where $L$ is free is something which looks like a presentation of a group by generators and relations ( the basis elements of $L$ are the generators, and the elements of $N$ are the relations), so we could say that the interest is the same: understanding $M$ by understanding the relations between the generators.
But really, one of the main interests is that is allows you to do computations. It is way easier to work when you have a basis.
Typically, take (as reuns emphasized) a finitely generated $R$-module $M$, so we have a surjective linear map $\pi:L_2\to M$, where $L_2$ is free of finite rank, so that $M\simeq L_2/\ker(\pi)$. If you can have a good understanding of $\ker(\pi)$, then you will be able to work in $M$ in an easier way.
In many situations, $\ker(\pi)$ is also finitely generated (for example, if $R$ is noetherian), so you can find a linear map $f:L_1\to L_2$, where $L_1$ is free of finite rank, whose image is $\ker(\pi)$. Namely, you will have an exact sequence $L_1\to L_2\to M\to 0$. And that's where it becomes interesting. One may show that $M$ is projective if and only there exists a linear map $g_0:L_2\to L_1$ satisfying $fg_0f=f$.
In this case, $L_2=\ker(g_0fg_0)\oplus \rm{im}(f)$ and $M\simeq \ker(g_0fg_0)$, so you have an explicit way to see $M$ as direct factor of a free module.
After choosing bases, using representative matrices, deciding whether or not $M$ is projective amounts to solve the inhomogeneous linear system $FG_0F=F$, where $F$ is the representative matrix $F$ of $f$ in the chosen bases.
I don't want to go into details, but in the case where $M$ is projective, the matrix $F$ allows you to compute explicitely elements $a_1,s_1,\ldots,a_r,s_r\in R$ such that $a_1s_1+\cdots+ars_r=1$ and $M_{s_i}$ is free of finite rank for all $i=1,\ldots,r$.
Let me continue with the case where $R$ is a PID. In this case, computing the Smith normal form of the matrix $F$ allows you to compute a basis of $L_2$ adapted to $\ker(\pi)$ and derive an isomorphism $M\simeq R^m\times R/a_1R\times \cdots\times R/a_s R$, where $a_i$ is non zero, non invertible and $a_1\mid \cdots\mid a_s$. This is the structure theorem for finitely generated modules over a PID, which has a lot of important applications (structure theorem for finitely generated abelian groups, Frobenius normal form of an endomorphism, which solves in an algorithmic way the similarity problem for endomorphisms...)
Finally, and veeeeery quickly, if you continue the process, for any module $M$ (finitely generated or not), and up to a change of notation, you can find an exact sequence $\cdots\to F_{n+1}\to F_n\to \cdots\to F_0\to M\to 0$, where each $F_j$ is free. That's what we call a free resolution of $M$.
Specialists of homological algebra will tell you that this kind of resolution (and more generally projective resolutions) are extremely important and useful.