When they solve $Av=-b$ the assumption is that $A$ is non-singular. Whenever $A$ is singular the problem can either have no solution, or infinitely many solutions. In their case I believe it is the latter. Remember that $v = (p,s)$. What they say is that they set the solution for $p$ to be the edge midpoint: $p=\frac{p_1+p_2}{2}$ and then plug that into the equation to reduce the system:
\begin{align}
-b = Av \iff -\begin{bmatrix} b_1\\ b_2\end{bmatrix} =
\begin{bmatrix}
A_{11} & A_{12}\\
A_{21} & I
\end{bmatrix}
\begin{bmatrix}
p \\ s \end{bmatrix} =
\begin{bmatrix} A_{11} \\ A_{21}\end{bmatrix}p+ \begin{bmatrix} A_{12} \\ I \end{bmatrix} s.
\end{align}
If you assume the system is consistent, then $s = -b_2 - A_{21}p$. If the system is inconsistent, then you can try to find the solution that least violates it as:
$$(A_{12}^TA_{12}+I)s = -A_{12}^T(b_1+A_{11}p) - (b_2+A_{21}p).$$
I suppose the system ought to be consistent though.