Write $S = R[x_1, \ldots, x_n]/J$ and denote $N = S \oplus M$ viewed as a module over $P = R[x_1, \ldots, x_n]$. If we can prove the lemma in case $S$ is a polynomial algebra over $R$, then we can find $I, a$ satisfying (1), (2), (3) such that the strict transform $N_{I, a}$ of $N$ along $R \to R[\frac{I}{a}]$ is flat over $R$ and finitely presented as a module over the strict transform $P_{I, a}]$ of $P$. Since $P_{I, a} = R[\frac{I}{a}][x_1, \ldots, x_n]$ (small detail omitted) we find that the summand $S_{I, a} \subset N_{I, a}$ is flat over $R$ and finitely presented as a module over $R[\frac{I}{a}][x_1, \ldots, x_n]$. Hence $S_{I, a}$ is finitely presented as an $R[\frac{I}{a}]$-algebra. Moreover, the summand $M_{I, a} \subset N_{I, a}$ is flat over $R$ and finitely presented as a module over $P_{I, a}$ hence also finitely presented as a module over $S_{I, a}$, see Algebra, Lemma \ref{algebra-lemma-finitely-presented-over-subring}. This reduces us to the case discussed in the next paragraph. \medskip\noindent Assume $S = R[x_1, \ldots, x_n]$. Choose a presentation $$ 0 \to K \to S^{\oplus r} \to M \to 0. $$ Let $M_A$ be the quotient of $M \otimes_R A$ by its torsion submodule, see Lemma \ref{lemma-torsion}. Then $M_A$ is a finite module over $S_A = A[x_1, \ldots, x_n]$. By Lemma \ref{lemma-valuation-ring-torsion-free-flat} we see that $M_A$ is flat over $A$. By Lemma \ref{lemma-flat-finite-type-valuation-ring-finite-presentation} we see that $M_A$ is finitely presented. Hence there exist finitely many elements $k_1, \ldots, k_t \in S_A^{\oplus r}$ which generate the kernel of the presentation $S_A^{\oplus r} \to M_A$ as an $S_A$-module. For any choice of $a \in I \subset \mathfrak m$ satisfying (1), (2), and (3) we denote $M_{I, a}$ the strict transform of $M$ along $R \to R[\frac{I}{a}]$. It is a finite module over $S_{I, a} = R[\frac{I}{a}][x_1, \ldots, x_n]$. By Algebra, Lemma \ref{algebra-lemma-valuation-ring-colimit-affine-blowups} we have $A = \colim_{I, a} R[\frac{I}{a}]$. This implies that $S_A = \colim S_{I, a}$ and $$ \colim M \otimes_R R[\textstyle{\frac{I}{a}}] = M \otimes_R A $$ Choose $I, a$ and lifts $k_1, \ldots, k_t \in S_{I, a}^{\oplus r}$. Since $M_A$ is the quotient of $M \otimes_R A$ by torsion, we see that the images of $k_1, \ldots, k_t$ in $M \otimes_R A$ are annihilated by a nonzero element $\alpha \in A$. After replacing $I, a$ by a different pair (recall that the colimit is filtered), we may assume $\alpha = x/a^n$ for some $x \in I^n$ nonzero. Then we find that $x k_1, \ldots, x k_t$ map to zero in $M \otimes_R A$. Hence after replacing $I, a$ by a different pair we may assume $x k_1, \ldots, x k_t$ map to zero in $M \otimes_R R[\frac{I}{a}]$ for some nonzero $x \in R$. Then finally replacing $I, a$ by $xI, xa$ we find that we may assume $k_1, \ldots, k_t$ map to $a$-power torsion elements of $M \otimes_R R[\frac{I}{a}]$. For any such pair $(I, a)$ we set $$ M'_{I, a} = S_{I, a}^{\oplus r}/ \sum S_{I, a}k_j. $$ Since $M_A = S_A^{\oplus r}/ \sum S_Ak_j$ we see that $M_A = \colim_{I, a} M'_{I, a}$. At this point we finally apply Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} (3) to conclude that $M'_{I, a}$ is flat for some pair $(I, a)$ as above. This lemma does not apply a priori to the system of strict transforms $$ M_{I, a} = (M \otimes_R R[\textstyle{\frac{I}{a}}])/a\text{-power torsion} $$ as the transition maps may not satisfy the assumptions of the lemma. But now, since flatness implies torsion free (Lemma \ref{lemma-flat-torsion-free}) and since $M_{I, a}$ is the quotient of $M'_{I, a}$ (because we arranged it so the elements $k_1, \ldots, k_t$ map to zero in $M_{I, a}$) by the $a$-power torsion submodule we also conclude that $M'_{I, a} = M_{I, a}$ for such a pair and we win.