Minimal free resolution

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.

I can't see that book online, but let me paraphrase the definition from Eisenbud's book on commutative algebra. See Chapter 19 page 473-477 for details.

Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m},$ then

Definition: A free resolution of a $R$-module $M$ is a complex

$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ... \rightarrow F_1 \rightarrow F_0$$

with trivial homology such that $\mathbf{\text{coker}}(F_1 \rightarrow F_0) \cong M$ and each $F_i$ is a free $R$-module.

He then defines a minimal resolution as follows:

Definition: A complex

$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ...$$

over $(R,\mathfrak{m})$ is minimal if the induced maps in the complex $\mathcal{F}\otimes R/\mathfrak{m}$ are each identically $0$. (Note that this is equivalent to the condition that $Im(F_i \rightarrow F_{i-1}) \subset \mathfrak{m}F_{i-1}$)

after this he proves the fact that a free resolution $\mathcal{F}$ is minimal if and only if a basis for $F_{i-1}$ maps into a minimal set of generators for $\mathbf{\text{coker}}(F_i \rightarrow F_{i-1}).$

The proof is via a straightforward appeal to Nakayama's Lemma.

From the way your question is worded, I would guess that your text has taken the latter of these equivalent conditions as the definition of a minimal free resolution and proved it to be equivalent to the former. But I can't be sure without seeing the text.