I am trying to show that the lyapunov equality:
$A^{T}P + PA = -Q$
Where $Q = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $
is equivalent to solving:
$M \begin{bmatrix} P_{11}\\ P_{12}\\ P_{22}\\ \end{bmatrix} = v $
The stated solution is:
$M:= \begin{bmatrix} -4 & -2 & 0\\ 1 & -2 & -1\\ 0 & 2 & 0\\ \end{bmatrix} $
and
$ v:= \begin{bmatrix} -1 \\ 0 \\ -1 \\ \end{bmatrix} $
I can see that the values for the vector $v$ come from the corresponding $Q_{11},Q_{12},Q_{22}$ values. However, I do not know how the M values are defined. If I multiply out the left side of the equation I get:
$ \begin{bmatrix} -4P_{11} - P_{12} - P_{21} & P_{11} - 2P_{12} - P_{22}\\ P_{11} - 2P_{21} - P_{22} & P_{12} + P_{21}\\ \end{bmatrix} $
Which if i try to map this to an M matrix I would get:
$M = \begin{bmatrix} -4 & -1 & 0\\ 1 & -2 & -1\\ 0 & 1 & 0\\ \end{bmatrix} $
Which is close, but not the same as the solution.
If in your case the Lyapunov equality can be written as
$$ \begin{bmatrix} -4\,P_{11} - P_{12} - P_{21} & P_{11} - 2\,P_{12} - P_{22} \\ P_{11} - 2\,P_{21} - P_{22} & P_{12} + P_{21} \end{bmatrix} = - \begin{bmatrix} Q_{11} & Q_{12} \\ Q_{21} & Q_{22} \end{bmatrix} \tag{1} $$
then by stacking the columns of $(1)$ on top of each other, then it can also be written as
$$ \begin{bmatrix} -4\,P_{11} - P_{12} - P_{21} \\ P_{11} - 2\,P_{21} - P_{22} \\ P_{11} - 2\,P_{12} - P_{22} \\ P_{12} + P_{21} \end{bmatrix} = - \begin{bmatrix} Q_{11} \\ Q_{21} \\ Q_{12} \\ Q_{22} \end{bmatrix}. \tag{2} $$
Factoring out $P$ gives
$$ \begin{bmatrix} -4 & -1 & -1 & 0 \\ 1 & -2 & 0 & -1 \\ 1 & 0 & -2 & -1 \\ 0 & 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} P_{11} \\ P_{21} \\ P_{12} \\ P_{22} \end{bmatrix} = - \begin{bmatrix} Q_{11} \\ Q_{21} \\ Q_{12} \\ Q_{22} \end{bmatrix}. \tag{3} $$
Now if we assume that $P = P^\top$ and $Q = Q^\top$, which implies $P_{21} = P_{12}$ and $Q_{21} = Q_{12}$, then the second and third equation from $(2)$ and $(3)$ are identical, so one of them can be omitted. Also the second column of the matrix, which gets multiplied by the $P$ vector, gets multiplied by $P_{21}$ which is identical to $P_{12}$. But the third column of the matrix gets multiplied by $P_{12}$. So this can be simplified down to that the sum of the second and third column get multiplied by $P_{12}$. Applying this to $(3)$ allows it to be rewritten as
$$ \begin{bmatrix} -4 & -2 & 0 \\ 1 & -2 & -1 \\ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} P_{11} \\ P_{12} \\ P_{22} \end{bmatrix} = - \begin{bmatrix} Q_{11} \\ Q_{12} \\ Q_{22} \end{bmatrix}. \tag{4} $$
So from this it can be seen that this indeed matches the stated solution for $M$. You probably just forgot to add the second and third column.
Also the matrix in the formulation in $(3)$ can easily be constructed for any $A \in \mathbb{R}^{2 \times 2}$ using
$$ \hat{M} = \begin{bmatrix} A + I\,A_{11} & I\,A_{12} \\ I\,A_{21} & A + I\,A_{22} \end{bmatrix}. \tag{5} $$
So from this it can also be deduced that your $A$ matrix should equal
$$ A = \begin{bmatrix} -2 & -1 \\ 1 & 0 \end{bmatrix}. $$
In general when $A \in \mathbb{R}^{n \times n}$ then $\hat{M} \in \mathbb{R}^{n^2 \times n^2}$, such that it consists out of a $n$ by $n$ grid of $\mathbb{R}^{n \times n}$ sub-matrices. By denoting $\hat{M}_{ij}$ as the sub-matrix on the $i$th place in the vertical direction and of the $j$th place in the horizontal direction, then each sub-matrix can be constructed using
$$ \hat{M}_{ij} = \left\{ \begin{array}{ll} A + A_{ij}\,I, & \text{if}\ i = j \\ A_{ij}\,I, & \text{otherwise} \end{array}\right. \tag{6} $$