$\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.
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 $M$.
