This answer is a slight reformulation of jspecter's; perhaps it will help.
A minimal free resolution is one in which each free module has the minimal number of generators. (Hence the name.)
Here is how you make it:
Start with $M$, f.g. over $R$. Its minimal number of generators is $\dim M/\mathfrak m M$. So we can take a free module $F_0$ with this number of generators and a surjection $F_0 \to M$ (but we can't do this with any
free module of smaller rank).
If this map is an isomorphism, we're done.
Otherwise, note that since $F_0/\mathfrak m F_0 \to M/\mathfrak m M$ is
a surjection between $R/\mathfrak m$-vector spaces of the same dimension,
it is an isomorphism, and so the kernel of the surjection $F_0 \to M$
is contained in $\mathfrak m F_0$. Now apply the same process that we
used to construct $F_0 \to M$ to this kernel,
to obtain a map of free modules $F_1 \to F_0$ whose cokernel is $M$,
now with both $F_0$ and $F_1$ being free on the minimal possible number
of generators.
The same argument as above will show that the kernel of $F_1 \to F_0$
is contained in $\mathfrak m F_1$.
We now continue inductively, and so produce a minimal free resolution of $M$.
As a biproduct of the construction, we find that each map $F_i \to F_{i-1}$ has image lying in $\mathfrak m F_{i-1}$. Equivalently, each map $F_i\to F_{i-1}$ reduces to the zero map modulo $\mathfrak m$.
Now, as jspecter points out,
there is a converse to the preceding remark: any free resolution with the property that $F_{i+1} \to F_{i}$ has image lying in $\mathfrak m F_{i}$ for every $i \geq 0$ (or equivalently, with the property that the maps $F_{i+1} \to F_{i}$ reduce to $0$ mod $\mathfrak m$ for $i \geq 0$) is a minimal free resolution, in the sense that the $i$th stage (for $i \geq 0$), the number of generators of $F_i$
is equal to the minimal number of generators of the kernel of the map
$F_{i-1} \to F_{i-2}$. (Here we agree that $F_{-1} = M$ and that $F_{-2} = 0$.)
(In jspecter's answer, things are phrased in terms of
of cokernels rather than kernels. But we are saying the same thing: since the $F_i$ form a complex,
the cokernel of $F_i \to F_{i-1}$ is the same as the kernel of $F_{i-1} \to F_{i-2}$. I have given a formulation in terms of kernels just because I find it slightly more intuitive.)
As jspecter also says,
in the book you are reading, the definition of minimal resolution is almost surely the one I give at the beginning of this answer. The book is then using the preceding remark and its converse to give the alternative characterization of minimal resolutions that you asked about.
The idea is the same used to get a basis for the dual space of a vector space $V$ of finite dimension.
Let us define the homomorphisms $g_i^*\colon G\to \mathbb{Z}$ such that $g_i^*(g_j):=\delta_{ij}$. Then given $f\in Hom(G,\mathbb{Z})$ you have that
$ f=\sum_{g_i}f(g_i)\cdot g_i^*$
Now suppose that $\sum_{g_i}a_ig_i^*=0$. Then $\left(\sum_{g_i}a_ig_i^*\right)(g_j)=a_j=0$ for each $j$. Hence $Hom(G,\mathbb{Z})$ is a free abelian group of rank $n$.
Best Answer
An abelian group is the same thing as a module over the ring $\mathbb{Z}$ (think about it). The ring $\mathbb{Z}$ is a PID, thus submodules of free $\mathbb{Z}$-modules are free. Reformulated in the context of abelian groups, submodules of free abelian groups are free abelian. You can use this fact to show that every abelian group has a length 2 free resolution (this works over any PID $R$, in particular $R = \mathbb{Z}$):
In the above proof, notice that "free $\mathbb{Z}$-module" is also the same thing as "free abelian group", so everything works out. You can also choose a set of generators for your module $M$, but I feel it's cleaner by just taking every element of $M$ and be done with it.
As Bernard mentions in the comments, for finitely generated abelian groups it's easy to see from the structure theorem (e.g. $\mathbb{Z}/n\mathbb{Z}$ has the free resolution $0 \to \mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$).
Other people have already commented on the difference between resolutions of abelian groups and groups in general. Surprisingly enough, subgroups of free groups are free too by the Nielsen–Schreier theorem, and the standard proof even uses algebraic topology! It all comes around. So you can directly adapt the above argument to show that every (not necessarily abelian) group has a free resolution of length at most two:
In the above proof, $\mathbb{Z}$ appears too, but for different reasons: it is the free group on one generator. It also happens to be the free abelian group on one generator, but that's not its role in the proof above. The groups $P_0$ and $P_1$ that appear in the proof are, in general, not free abelian.