$\let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} $ I will write $R$ for $R_{q}$, because $q$ is constant. In the following, I am going to assume that $R$ is an arbitrary right inverse of $M_{q}$, rather than the specific right inverse $M_{q}^{T}\left( M_{q}M_{q}^{T}\right) ^{-1}$ which you have suggested. This makes the statement a tad more general and rids us of a red herring.
I shall prove that for any $i\in\left\{ 1,2,...,n\right\} $, $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, the value $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (independently, and independent, of the choice of right inverse $R$).
For every $\ell\in\left\{ 1,2,...,n\right\} $, we abbreviate $C_{q\ell}$ by $v_{q}$. Thus,
$v_{\ell}=C_{q\ell}=\left( -1\right) ^{q+\ell}\det\left( \underbrace{M\text{ without row }q}_{=M_{q}}\text{ and column }\ell\right) $
(1) $=\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $.
From this it easy to obtain
(2) $\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}=0$ for every $j \in \left\{1,2,...,n-1\right\}$.
[Proof of (2): Let $j \in \left\{1,2,...,n-1\right\}$. Let $G$ be the result of inserting a copy of row $j$ of the matrix $M_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields
$\det G=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $,
since the entries of this row are $\left( M_{q}\right) _{j\ell}$ and the cofactors corresponding to this row are $\left( -1\right) ^{q+\ell} \det\left( M_{q}\text{ without column }\ell\right) $. Now, (1) yields
$\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }v_{\ell}$
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column } \ell\right) $
$=\det G=0$,
so that (2) is proven.]
Since $M$ is generic, its cofactors $v_{1}$, $v_{2}$, $...$, $v_{n}$ are nonzero.
Fix $i\in\left\{ 1,2,...,n\right\} $. Let $N_{i}$ denote the matrix $M_{q}$ without column $i$. Thus, $\det\left( N_{i}\right) =\det\left( M_{q}\text{ without column }i\right) =\left( -1\right) ^{q+i}v_{i}$ (because (1) yields $v_{i}=\left( -1\right) ^{q+i}\det\left( M_{q}\text{ without column }i\right) $).
Let $S_{i}$ denote the matrix $\left( R_{jk}-\dfrac{v_{j}}{v_{i}} R_{ik}\right) _{j\in\left\{ 1,2,...,n\right\} ;\ k\in\left\{ 1,2,...,n-1\right\} }$. The $i$-th row of this matrix $S_{i}$ is zero; let $S_{i}^{\prime}$ denote the matrix $S_{i}$ without row $i$.
Now, we claim that $N_{i}S_{i}^{\prime}=I_{n-1}$ (the $\left( n-1\right) \times\left( n-1\right) $ identity matrix). Indeed, for every $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$, the $\left( j,k\right) $-th entry of $N_{i}S_{i}^{\prime}$ is
$\left( N_{i}S_{i}^{\prime}\right) _{jk}=\sum_{u\in\left\{ 1,2,...,n-1\right\} }\left( N_{i}\right) _{ju}\left( S_{i}^{\prime }\right) _{uk}$
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\underbrace{\left( S_{i}\right) _{\ell k} }_{=R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}}$
(since $N_{i}$ is the matrix $M_{q}$ without column $i$, while $S_{i}^{\prime }$ is the matrix $S_{i}$ without row $i$)
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i} }R_{ik}\right) $
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}\right) $
(here, we added an $\ell=i$ addend to the sum; this did not change the sum because this addend is $0$)
$=\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}R_{\ell k}}_{\substack{=\left( M_{q}R\right) _{jk}=\delta _{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M_{q}\text{)} }}-\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\dfrac{v_{\ell}}{v_{i}}R_{ik}}_{=\dfrac{1}{v_{i}}R_{ik}\sum_{\ell \in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}}$
$=\delta_{jk}-\dfrac{1}{v_{i}}R_{ik}\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}} _{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta_{jk}$.
This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime} =N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left( n-1\right) \times\left( n-1\right) $-matrices).
Now, we recall Cramer's rule for the inverse of a matrix. It essentially says that the inverse of a square matrix is obtained by dividing its adjoint by its determinant. In other words, if $X$ is an invertible $p\times p$-matrix for some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{ 1,2,...,p\right\} $, then
(3) $\left( X^{-1}\right) _{jk}=\dfrac{1}{\det X}\left( -1\right) ^{j+k}\det\left( X\text{ without row }k\text{ and column }j\right) $.
Now, recall that $S_{i}^{\prime}=N_{i}^{-1}$. Hence,
$\left( S_{i}^{\prime}\right) _{jk}=\left( N_{i}^{-1}\right) _{jk} =\dfrac{1}{\det\left( N_{i}\right) }\left( -1\right) ^{j+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (by (3))
$=\dfrac{1}{\left( -1\right) ^{q+i}v_{i}}\left( -1\right) ^{j+k} \det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (since $\det\left( N_{i}\right) =\left( -1\right) ^{q+i}v_{i}$)
$=\dfrac{1}{v_{i}}\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $
for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$. In other words,
(4) $v_{i}\left( S_{i}^{\prime}\right) _{jk}=\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $
for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$.
Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ to $\left\{ 1,2,...,n-1\right\} $. Then, for all $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, we have
(5) $v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B} \left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $.
(Indeed, this follows from (4), applied to $\mathbf{B}\left( j\right) $ instead of $j$, because $\left( S_{i}^{\prime}\right) _{\mathbf{B}\left( j\right) ,k}=\left( S_{i}\right) _{jk}$ (since $S_{i}^{\prime}$ is the matrix $S_{i}$ without row $i$) and because column $j$ of $M_{q}$ is column $\mathbf{B}\left( j\right) $ of $N_{i}$ (since $N_{i}$ is the matrix $M_{q}$ without column $i$).)
Now, fix $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $. We need to show that $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$. We WLOG assume that $j\neq i$ (since otherwise, $C_{qi}R_{jk}-C_{qj}R_{ik} =0$). Thus, $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $. Since $C_{qi}=v_{i}$ and $C_{qj}=v_{j}$ (by the definitions of $v_{i}$ and $v_{j}$), we have
$\underbrace{C_{qi}}_{=v_{i}}R_{jk}-\underbrace{C_{qj}}_{=v_{j}}R_{ik} =v_{i}R_{jk}-v_{j}R_{ik}=v_{i}\underbrace{\left( R_{jk}-\dfrac{v_{j}}{v_{i} }R_{ik}\right) }_{\substack{=\left( S_{i}\right) _{jk}\\\text{(by the definition of }S_{i}\text{)}}}$
(6) $=v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B}\left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $
(by (5)),
which is obviously a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (and independent of the choice of $R$), qed.
Thanks for a very nice question, and sorry for this mess of an answer...
PS. I believe the genericity of $M$ is not required for (6) to hold. Does anyone see a nice proof of this? I don't.
PPS. Here is a more general statement which, I think, is true. Let $A$ be a commutative ring. Let $N$ be an $\left(n-1\right) \times n$-matrix over $A$. Let $s \in A^n$ be a vector. For every $\ell \in \left\{1,2,...,n\right\}$, let $p_\ell$ denote the scalar $\left( -1\right) ^{\ell}\det\left( N \text{ without column }\ell\right)$. If $w$ is a vector and $i$ is an integer, we denote by $w_i$ the $i$-th coordinate of $w$ (whenever this makes sense). Then, every two distinct $i \in \left\{1,2,...,n\right\}$ and $j \in \left\{1,2,...,n\right\}$ satisfy
(11) $p_i s_j - p_j s_i = \sum_{k=1}^{n-1} \pm \left(Ns\right)_k \det\left(N \text{ without row } k \text{ and columns } i \text{ and } j\right)$,
where $\pm$ is something like $\left(-1\right)^{i+j+\left[i<j\right]+k}$ using the Iverson bracket.
If this is proven (and this shouldn't be too hard -- it's a polynomial identity, so you can assume as much genericity as you wish), the original result is obtained by setting $N = M_q$ and $s = \left(k\text{-th column of } R\right)$.
PPPS. Indeed, (11) is not hard to prove. Let $e_1, e_2, ..., e_n$ be the $n$ standard basis vectors of $A^n$. Let $P$ be the $n \times \left(n-1\right)$-matrix whose columns (from left to right) are $e_1, e_2, ..., \widehat{e_i}, ..., \widehat{e_j}, ..., e_n, s$ (where $\widehat{\text{something}}$ means omission, and the order of $i$ and $j$ is not necessarily the one we have shown). Then, $NP$ is the $\left(n-1\right)\times \left(n-1\right)$-matrix whose columns (from left to right) are the columns $1, 2, ..., \widehat{i}, ..., \widehat{j}, ..., n$ of $N$ and the column-vector $Ns$. The right hand side of (11) is the determinant of this matrix $NP$, computed by Laplace expansion along its last column. The left hand side of (11) is the same determinant, computed using the Cauchy-Binet formula (which takes a rather simple form here because if we remove the $\ell$-th row from $P$ for some $\ell \notin \left\{i, j\right\}$, then the resulting matrix has two rows with only their rightmost entries nonzero, and so has determinant $0$).
PPPPS. Here is a different way to formulate the above proof of (11), using exterior algebra instead of the Cauchy-Binet formula and giving some more detail.
Let $e_{1}$, $e_{2}$, $...$, $e_{n}$ be the $n$ standard basis vectors of $A^{n}$. Let $f_{1}$, $f_{2}$, $...$, $f_{n-1}$ be the $n-1$ standard basis vectors of $A^{n-1}$. We identify the matrix $N\in A^{\left( n-1\right) \times n}$ with the $A$-linear map $A^{n}\rightarrow A^{n-1}$ it represents. As usual, we use the notation ''$\widehat{\text{some term}}$'' for omission of a term in a product or list (for example, $\left( 1,2,...,\widehat{5},...,8\right) =\left( 1,2,3,4,6,7,8\right) $). We also use the Iverson bracket notation, i.e., whenever $\mathcal{A}$ is a logical statement, we write $\left[ \mathcal{A}\right] $ for the integer $\left\{ \begin{array} [c]{c} 1,\text{ if }\mathcal{A}\text{ is true;}\\ 0,\text{ if }\mathcal{A}\text{ is false} \end{array} \right. $.
Let $\mathbf{f}$ be the element $f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}$ of $\wedge^{n-1}\left( A^{n-1}\right) $. It is known that $\left( \mathbf{f}\right) $ is an $A$-module basis of $\wedge^{n-1}\left( A^{n-1}\right) $.
Let $i$ and $j$ be two distinct elements of $\left\{ 1,2,...,n\right\} $. We have $s=\sum_{k=1}^{n}s_{k}e_{k}$, so that
$e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j} }\wedge...\wedge e_{n}\wedge s$
$=e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge \widehat{e_{j}}\wedge...\wedge e_{n}\wedge\left( \sum_{k=1}^{n}s_{k} e_{k}\right) $
(12) $=\sum_{k=1}^{n}s_{k}e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i} }\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge e_{k}$
(where the notation $e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}} \wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}$ means that both $e_{i}$ and $e_{j}$ are omitted, but does not necessarily imply that $i<j$). Of all the addends of the sum on the right hand side of (12), only those for $k=i$ and for $k=j$ have a chance to be nonzero (every other summand contains a wedge product with two equal factors, and thus vanishes). Hence, (12) simplifies to
$e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j} }\wedge...\wedge e_{n}\wedge s$
$=s_{i}\underbrace{e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}} \wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge e_{i}}_{=\left( -1\right) ^{i+\left[ i<j\right] }e_{1}\wedge e_{2}\wedge...\wedge \widehat{e_{j}}\wedge...\wedge e_{n}}+s_{j}\underbrace{e_{1}\wedge e_{2} \wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge e_{j}}_{=\left( -1\right) ^{j+1-\left[ i<j\right] }e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge e_{n}}$
$=s_{i}\left( -1\right) ^{i+\left[ i<j\right] }e_{1}\wedge e_{2} \wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}+s_{j}\left( -1\right) ^{j+1-\left[ i<j\right] }e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i} }\wedge...\wedge e_{n}$
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s_{j}\left( -1\right) ^{i}e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge e_{n}-s_{i}\left( -1\right) ^{j}e_{1}\wedge e_{2}\wedge...\wedge \widehat{e_{j}}\wedge...\wedge e_{n}\right) $.
Applying the linear map $\wedge^{n-1}N$ to both sides of this equality results in
$\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge \widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right) $
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s_{j}\left( -1\right) ^{i}\underbrace{\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i}}\wedge...\wedge e_{n}\right) } _{=\det\left( N\text{ without column }i\right) \mathbf{f}}\right. $
$\left. -s_{i}\left( -1\right) ^{j}\underbrace{\left( \wedge ^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{j}} \wedge...\wedge e_{n}\right) }_{=\det\left( N\text{ without column }j\right) \mathbf{f}}\right) $
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s_{j} \underbrace{\left( -1\right) ^{i}\det\left( N\text{ without column }i\right) }_{=p_{i}} \mathbf{f} -s_{i}\underbrace{\left( -1\right) ^{j} \det\left( N\text{ without column }j\right) }_{=p_{j}} \mathbf{f} \right) $
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s_{j}p_{i} \mathbf{f}-s_{i}p_{j}\mathbf{f}\right) =\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( p_{i}s_{j}-p_{j}s_{i}\right) \mathbf{f}$,
so that
$\left( p_{i}s_{j}-p_{j}s_{i}\right) \mathbf{f}$
(13) $=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i} }\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right) $.
On the other hand, using $\wedge$ to denote the multiplication in the exterior algebra $\wedge\left( A^{n-1}\right) $, we see
$\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge \widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right) $
$=\underbrace{\left( \wedge^{n-2}N\right) \left( e_{1}\wedge e_{2} \wedge...\wedge\widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\right) }_{=\sum_{\ell=1}^{n-1}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) f_{1}\wedge f_{2}\wedge ...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}}\wedge\underbrace{\left( Ns\right) }_{=\sum_{k=1}^{n-1}\left( Ns\right) _{k}f_{k}}$
$=\left( \sum_{\ell=1}^{n-1}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) f_{1}\wedge f_{2}\wedge...\wedge \widehat{f_{\ell}}\wedge...\wedge f_{n-1}\right) \wedge\left( \sum _{k=1}^{n-1}\left( Ns\right) _{k}f_{k}\right) $
(14) $=\sum_{\ell=1}^{n-1}\sum_{k=1}^{n-1}\left( Ns\right) _{k} \det\left( N\text{ without columns }i\text{ and }j\text{ and row } \ell\right) f_{1}\wedge f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}\wedge f_{k}$.
Of all the addends of the inner sum on the right hand side of (14), only those for $k=\ell$ have a chance to be nonzero (because all other addends have the form $f_{1}\wedge f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}\wedge f_{k}$ for $k\neq\ell$, which is a wedge product with two equal factors and thus equals $0$). Hence, (14) simplifies to
$\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge \widehat{e_{i}}\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right) $
$=\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \underbrace{f_{1}\wedge f_{2}\wedge...\wedge\widehat{f_{\ell}}\wedge...\wedge f_{n-1}\wedge f_{\ell} }_{=\left( -1\right) ^{n-1-\ell}f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}}$
$=\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right) ^{n-1-\ell}\underbrace{f_{1}\wedge f_{2}\wedge...\wedge f_{n-1}}_{=\mathbf{f} }$
$=\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right) ^{n-1-\ell}\mathbf{f}$.
Thus, (13) becomes
$\left( p_{i}s_{j}-p_{j}s_{i}\right) \mathbf{f}$
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\underbrace{\left( \wedge^{n-1}N\right) \left( e_{1}\wedge e_{2}\wedge...\wedge\widehat{e_{i} }\wedge...\wedge\widehat{e_{j}}\wedge...\wedge e_{n}\wedge s\right) } _{=\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right) ^{n-1-\ell}\mathbf{f}}$
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right) ^{n-1-\ell}\mathbf{f}$.
Since $\left( \mathbf{f}\right) $ is a basis of $\wedge^{n-1}\left( A^{n-1}\right) $, we can compare coefficients before $\mathbf{f}$ in this equality, and obtain
$p_{i}s_{j}-p_{j}s_{i}$
$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\sum_{\ell=1}^{n-1}\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right) ^{n-1-\ell}$
$=\sum_{\ell=1}^{n-1}\left( -1\right) ^{\left[ i<j\right] +i+j+n-\ell }\left( Ns\right) _{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }\ell\right) $
$=\sum_{k=1}^{n-1}\left( -1\right) ^{\left[ i<j\right] +i+j+n-k}\left( Ns\right) _{k}\det\left( N\text{ without columns }i\text{ and }j\text{ and row }k\right) $.
This proves (11), up to all the sign errors I surely have made.
As a consequence, we obtain a new proof of your statement, and it does no longer rely on genericity of $M$.