As you know, Nullstellensatz implies that $m= (x_1 -c_1, \dots, x_n-c_n)$. That means $m$ is generated by the $\lbrace x_i-c_i\rbrace$, and thus, $m$ is the ideal of all polynomials $p(x_1,\dots,x_n)$ such that $p(c_1,\dots,c_n)= 0$.
The residue field $R/m$ is isomorphic to $K$ (recall that the map $R\rightarrow K$ given by $p(x_1,\dots,x_n)\mapsto p(c_1,\dots,c_n)$ is surjective with kernel $m$). so you have to show that $m/m^2$ is $n$-dimensional as a $K$-vector space.
We are going to build a basis for this space, with $n$ elements.
Consider, for every $p(x_1,\dots,x_n)\in m$, the Taylor expansion at $(c_1, \dots, c_n)$:
$$p(x_1,\dots,x_n)= \sum_{i=1}^n \frac{\partial p}{\partial x_i}(c_1,\dots,c_n)(x_i-c_i) + R(x_1,\dots x_n) $$
With $R\in m^2$. Taking residues modulo $m^2$ shows that the class of $p$ in $m/m^2$ is a linear combination of the classes of $x_i-c_i$ (with coefficients equal to the partial derivatives), which are therefore a system of generators of $m/m^2$.
Nakayama's Lemma then implies that the classes of $x_i-c_i$ in $R_m$ are a system of generators of $mR_m$.
Still, to see that they are linearly independent, if there was a linear combination equal to $0$:
$$a_1(x_1-c_1) + \dots a_n(x_n-c_n) = 0\quad mod.\ m/m^2$$
Then $a_1(x_1-c_1) + \dots a_n(x_n-c_n)\in m^2$, so its first partial derivatives are zero at $(c_1, \dots, c_n)$, but they are equal to the $a_i$.
Best Answer
$$\frac{R[X]}{mR[X]+(X)}\simeq\frac{R[X]/(X)}{mR[X]+(X)/(X)}\simeq R/m$$
$R[X]_N$ is regular since $R[X]$ is regular and localizations of regular rings are regular. (Alternatively, notice that $X\notin (NR[X]_N)^2$, and $R[X]_N/XR[X]_N\simeq R_m=R$ is regular.)