Let $M$ be an $A$-module for some commutative ring $A$ with unit. Define a free resolution for $M$ to be an exact sequence of free modules $$\cdots \longrightarrow N_2 \longrightarrow N_1 \longrightarrow N_0 \longrightarrow M \longrightarrow 0.$$
In Hatcher's algebraic topology, Hatcher claims that if $M$ is an abelian group, then $M$ has a free resolution of the form $0 \longrightarrow N_1 \longrightarrow N_0 \longrightarrow M \longrightarrow 0$. I can't seem to find a reference for an analogous result for arbitrary modules. My main question is: does every module over a commutative ring admit a free resolution? Not necessarily finte.
Secondly, consider that for a sheaf $\mathscr{S}$ of $\mathscr{R}$-modules, we define a free resolution of $\mathscr{R}$-modules to be an exact sequence of sheaves $$\cdots \longrightarrow \mathscr{R}^2 \longrightarrow \mathscr{R}^1 \longrightarrow \mathscr{R} \longrightarrow \mathscr{S} \longrightarrow 0,$$ where $\mathscr{R}^k$ denotes the sheaf of free modules given by the $k$th direct sum of $\mathscr{R}$.
It is remarked in a few texts in algebraic geometry that the admission of a free resolution of $\mathscr{R}$-modules is a global and rather stringent condition. This motivates the definition of a coherent sheaf of $\mathscr{R}$-modules.
Can someone elaborate on why this free resolution is not induced by the corresponding sequence on stalks? Also, why is this condition much more elaborate for sheaves than modules? Can someone provide an illustrative example to show the difficulty in constructing a free resolution for a sheaf of modules?
It is rather confusing that you are dealing with sheaves without knowing that categories of modules always have enough projectives, because they have enough free modules: if $R$ is a ring (ommutative or not) and if $M$ is any $R$-module, let $FM$ be the free $R$-module with basis the set $M$. There is a surjection $e_M : FM \to M$ that assigns $e_m\to m$. Let $K$ be the kernel of the map $e_M$, and let $e_K: FK\to K$ be as before. This gives an exact sequence $FK\to FM\to M$. Continue in this way to obtain a functorial free resolution of $M$.