$\newcommand{\ideal}[1]{\mathfrak{#1}}\newcommand{\tensor}{\otimes}$
The correct condition for $J$ in the case $m < n$ is, that if $J$ is the ideal of the $m \times m$ minors of $M = (\partial f_i / \partial T_j)_{ij}$ and $I=(f_1,\ldots,f_m)$ then $J + I = R[T_1,\ldots,T_n] = A$.
If $B = A/I$ we want to show for all maximal ideals $\ideal{n} \subseteq A$ and $\ideal{p} = \ideal{n} \cap R$ the extension $B_\ideal{n}/R_\ideal{p}$ is flat.
For $\ideal{n} \not\supseteq I$ this is true, as $B_\ideal{n} = 0$. So only the $\ideal{n} \supseteq I$ remain.
Calling $k = k(\ideal{p}) = R_\ideal{p}/\ideal{p} R_\ideal{p}$ we have to consider the rings $B_p = A/(f_1,\ldots,f_p)$ and their base changes
$C_p = (B_p)_\ideal{n} \tensor_R k(\ideal{p}) = (k[T_1,\ldots,T_n]/(\bar{f}_1,\ldots,\bar{f}_p))_\ideal{n}$ where $\bar{f}$ is the image in $A \tensor_R k$ of $f \in A$.
Now, the key to the further argument is the observation, that because of the condition on the minors of $M$, the polynomials $\bar{f}_1,\ldots,\bar{f_p}$ define a nonsingular variety in $\mathrm{spec}(k[T_1,\ldots,T_n])$ (the Jacobian $(\partial f_i/\partial T_j)_{{i=1,\ldots,p}\atop {j=1,\ldots,n}} \tensor_R k$ has full rank $p$ at $\ideal{n}$). So $C_p$ is an integral domain and $\bar{f}_{p+1}$ is a non-zero divisor in $C_p$ because it is not contained in the ideal $(\bar{f}_1,\ldots,\bar{f}_p)$.
So by the proposition 2.5 cited, as $(B_0)_\ideal{n} = A_\ideal{n}$ is $R_\ideal{p}$-flat, so is $(B_1)_\ideal{n} = (B_0)_\ideal{n}/(f_1)$ and finally, by induction also $(B_m)_\ideal{n} = B_\ideal{n}$.
The correct version of proposition 2.5 is:
Let $B$ be a flat $A$-algebra and consider $b \in B$. If the image of $b$ in $B \tensor_A k(\ideal{n} \cap A)$ for all $\ideal{n} \subseteq B$, maximal, is a non-zero-divisor, then $B/bB$ is a flat $A$-algebra.
Yes. The proof is not really accurate. For any $y' \in \mathbb{A}^{n}_{\kappa(y)}$ lying over $y\in \mathbb{A}^{n}_k$, it is not in general $\kappa(y') =\kappa(y)$, so we cannot apply the reduced case directly. But we can choose suitable $y'$ such that $\kappa(y') =\kappa(y)$. This is as follows.
Again let $f: X \to \operatorname{Spec} k$ , $l : \operatorname{Spec} \kappa(y) \to \operatorname{Spec}k$, and $i_y : \operatorname{Spec}\kappa(y) \to X$ be the canonical morphism ( Gortz's book p.71 ). Since $ l \circ id_{\operatorname{Spec} \kappa(y)} =l = \operatorname{Spec}( k \hookrightarrow \kappa(y)) := \operatorname{Spec}(\Gamma(f \circ i_y)) = f\circ i_y $, by universal property, there is an unique morphism $g:=\operatorname{Spec}\kappa(y) \to \overline{X}:=X \otimes_k \kappa(y)$ such that $p\circ g = i_y$ and $\overline{f} \circ g = id_{\operatorname{Spec}\kappa(y)} $. ($p$ and $\overline{f}$ are as in the commutative diagram in the question. ) And let $y'$ be its image point. Then $y'$ is lying over $y$. And since $g: \operatorname{Spec}\kappa(y) \to \overline{X}:=X \otimes_k \kappa(y)$ is a morphism of locally ringed spaces, it induces a local homomorphism $\mathcal{O}_{\overline{X},y'}\to \kappa(y)=\mathcal{O}_{\operatorname{Spec}\kappa(y),p}$, and hence a homomorphism $\iota :\kappa(y') \to \kappa(y)$. And also, we have field homomorphism $\eta : =\Gamma(\overline{f} \circ i_{y'}) : \kappa(y) \to \kappa(y')$, where $\overline{f} : \overline{X} \to \operatorname{Spec}\kappa(y)$ is the base change map in the question. Note that $\iota \circ \eta = id_{\kappa(y)}$ (?) so that $\iota$ is surjective hence an isomorphism ; i.e., $\kappa(y') \cong \kappa(y)$.
So we may apply the reduced case so we can understand the bold statement in the question.
P.s. I would like to thank Gortz for his advice through E-mail.
Best Answer
In $\frac{M}{M^2}$, assume $\exists \bar{g_i}\in \frac{R}{M}=\kappa(y)$ s.t. $\sum_i \bar{g_i}\cdot f_i=0$. So $\sum_i g_if_i\in M^2$.
For each $j$, we have $\frac{\partial \sum g_if_i}{\partial T_j}=\sum_i (\frac{\partial g_i}{\partial T_j}f_i+g_i\frac{\partial f_i}{\partial T_j})$.
Note that $\frac{\partial \ \cdot}{\partial T_j}$ sends elements in $M^2$ to $M$, plus that $f_i\in M$, we have $$\sum_i g_i\frac{\partial f_i}{\partial T_j}\in M$$ View it in $\frac{R}{M}=\kappa(y)$, it is zero.
By assumption $J(y):=J_{f_1,...,f_r}(y)$ as a matrix over $\kappa(y)$ has rank $r$, so for any $(\bar{g_1},...,\bar{g_r})\in \kappa(y)^r$, if $(\bar{g_1},...,\bar{g_r})J=0$, then $(\bar{g_1},...,\bar{g_r})=0$.
But $(\bar{g_1},...,\bar{g_r})J=0$ (in $\kappa(y)$) is exactly $\sum_i g_i\frac{\partial f_i}{\partial T_j}\in M,\forall j$. So we are done.