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.
Since $R$ is regular, there is the Koszul complex $K^\bullet$ resolving $k$. Assume $\mathfrak{m} = (x_1, \ldots, x_d)$ where $d = \dim R$.
The homology of $K^\bullet \otimes M$ is $\mathrm{Tor}^R_i(k, M)$. On the other hand, the last homology is the kernel of the map
$M \to M^d$ according to $m \mapsto (x_1m, \ldots, x_d m)$
which is clearly isomorphic to $\mathrm{Hom}(k, M)$. With $M = R/I$, this shows that the dimension of the socle is the top Betti number $\beta_d(R/I)$.
Best Answer
1) Your question is somewhat broad, but in some sense the general answer is "they don't", or at least that the relationship is not well understood in general, and I think constructing a minimal graded free resolution of $\operatorname{gr}_{\mathfrak{m}}(M)$ is quite a hard problem. Given a minimal free resolution $F_{\bullet}$ of $M$ over a local ring, one can construct an associated graded complex $\operatorname{gr}_{\mathfrak{m}} (F_{\bullet})$ of free modules, but it is rarely acyclic. In fact, $\operatorname{gr}_{\mathfrak{m}}(F_{\bullet})$ is a minimal graded free resolution of $\operatorname{gr}_{\mathfrak{m}}(M)$ if and only if $M$ is Koszul (this is sometimes taken as the definition of Koszul). See this paper for a reference on Koszul modules.
2) No. For example, let $S=k[\![x,y,z]\!]$ with $\mathfrak{n}=(x,y,z)$ and consider the $S$-module $R=S/I$, where $I$ is the ideal $I=(xz-y^3,yz-x^4,z^2-x^3y^2)$; one can prove $R \cong k[\![t^4,t^5,t^{11}]\!]$. This is a classical example of a domain of dimension $1$ whose associated graded ring is not Cohen-Macaulay, and it can be found, for instance, as Example 1.3 in these notes. Concretely, one can show that $\operatorname{gr}_{\mathfrak{n}}(R) \cong k[x,y,z]/(xz,yz,z^2,y^4)$. We observe that $z$ is a nonzero socle element of $\operatorname{gr}_{\mathfrak{n}}(R)$, so $\operatorname{gr}_{\mathfrak{n}}(R)$ has depth $0$. In particular, the Auslander-Buchsbaum formula tells us that the minimal graded $\operatorname{gr}_{\mathfrak{n}}(S)$-free resolution of $\operatorname{gr}_{\mathfrak{n}}(R)$ does not even have the same length as the minimal $S$-free resolution of $M$. That is, these modules have different projective dimensions.
The takeaway from these things is that the associated graded operation does not behave well with taking homology or with homological properties.