Following on from the matrix inversion lemma above, I think I have found an answer to my own question. To recap, the Schur complement is traditionally used for matrix inversion. If $A$ is invertible then it is a well-known property of the Schur complement that:
$$
M^{-1} = \begin{bmatrix}
A^{-1}+A^{-1}B(M/A)^{-1}CA^{-1} & -A^{-1}B(M/A)^{-1} \\
-(M/A)^{-1}CA^{-1} & (M/A)^{-1}
\end{bmatrix}
$$
And if $D$ is invertible then, similarly:
$$
M^{-1} = \begin{bmatrix}
(M/D)^{-1} & -(M/D)^{-1}BD^{-1} \\
-D^{-1}C(M/D)^{-1} & D^{-1} + D^{-1}C(M/D)^{-1}BD^{-1}
\end{bmatrix}
$$
Since the matrix inverse is unique, we can set these expressions for $M^{-1}$ equal to each other to obtain the following expressions, which are commonly known as the matrix inversion lemma:
$$
\begin{gathered}
(M/A)^{-1} = D^{-1} + D^{-1}C(M/D)^{-1}BD^{-1} \\
(M/D)^{-1} = A^{-1} + A^{-1}B(M/A)^{-1}CA^{-1}
\end{gathered}
$$
Focusing on $(M/A)^{-1}$ we can left- and right-multiply by $D$, then rearrange to obtain the following. $I_n$ is the $n\times n$ identity matrix where $D$ is $n\times n$.
$$
\begin{gathered}
D(M/A)^{-1}D = DD^{-1}D + DD^{-1}C(M/D)^{-1}BD^{-1}D \\
D(M/A)^{-1}D = D + C(M/D)^{-1}B \\
0 = C(M/D)^{-1}B + D - D(M/A)^{-1}D \\
0 = C(M/D)^{-1}B + \frac{1}{2}I_nD + D\left(\frac{1}{2}I_n\right) - D(M/A)^{-1}D \\
0 = C(M/D)^{-1}B + \frac{1}{2}I_nD + D\left(\frac{1}{2}I_n\right)^\top - D(M/A)^{-1}D
\end{gathered}
$$
James Potter describes a method for obtaining matrix quadratic solutions. The above can be rewritten as:
$$
0 = \mathcal{A} + \mathcal{B}X + X\mathcal{B}^\top - X\mathcal{C}X \\
$$
Where:
$$
\begin{gathered}
\underset{n\times n}{X} = D \\
\mathcal{A} = C(M/D)^{-1}B \\
\mathcal{B} = \frac{1}{2}I_n \\
\mathcal{C} = (M/A)^{-1}
\end{gathered}
$$
Then we let:
$$
\underset{2n\times 2n}{\mathcal{M}} = \begin{bmatrix}\mathcal{B} & \mathcal{A} \\ \mathcal{C} & -\mathcal{B}^\top\end{bmatrix} = \begin{bmatrix}\frac{1}{2}I_n & C(M/D)^{-1}B \\ (M/A)^{-1} & -\frac{1}{2}I_n\end{bmatrix}
$$
Now we can obtain the solution via the eigendecomposition of $\mathcal{M}$. Let $\mathcal{M}$ have eigenvectors $v_i$ where each of $v_i$ is a column vector with dimensions $2n\times 1$. We can divide each column vector $v_i$ into an upper half $u_i$ and lower halve $l_i$.
$$
v_i = \begin{bmatrix} v_1 \\ \vdots \\ v_{n} \\ v_{n+1} \\ \vdots \\ v_{2n} \end{bmatrix} = \begin{bmatrix} u_1 \\ \vdots \\ u_{n} \\ l_1 \\ \vdots \\ l_{n} \end{bmatrix}
$$
Then the solution will have the form:
$$
\underset{n\times n}{X} = \begin{bmatrix}u_1 & \cdots & u_n\end{bmatrix}\begin{bmatrix}l_1 & \cdots & l_n\end{bmatrix}^{-1}
$$
Potter's paper goes into further details about which $n$ eigenvectors to use. In my case, the original $M$ is symmetric positive definite, which implies that $A$ and $D$ are also SPD, and so the solution $X$ must be SPD. Under this assumption, we pick the $n$ eigenvectors $v_i$ whose eigenvalues are positive.
Note also that, if $M$ is symmetric, we can simplify the expressions for $\mathcal{M}$ by substituting $C=B^\top$.
For further intuition, consider the case where $D=d$ is scalar, in which case $B=b$ and $C=c$ are column and row vectors, respectively. Then we would have:
$$
D(M/A)^{-1}D - D - C(M/D)^{-1}B = d^2(M/A)^{-1} - d - c(M/D)^{-1}b = 0
$$
Applying the quadratic equation:
$$
\begin{gathered}
d = \frac{-(-1) \pm \sqrt{(-1)^2-4\left[(M/A)^{-1}\right]\left[-c(M/D)^{-1}b\right]}}{2(M/A)^{-1}} \\
d = \frac{1 \pm \sqrt{(1+4(M/A)^{-1}c(M/D)^{-1}b}}{2(M/A)^{-1}}
\end{gathered}
$$
We obtain two solutions for $d$ via the $\pm$ and must carefully choose whether to add or subtract. This is analagous to the matrix quadratic solution, in which we have a degenerate choice of which eigenvectors to use. In the case that $M$ is symmetric positive definite, then we know that $d > 0$, $(M/A)>0$, $(M/D)$ is SPD, and we can substitute $c=b^\top$ to obtain:
$$
d = \frac{1 + \sqrt{(1+4(M/A)^{-1}b^\top(M/D)^{-1}b}}{2(M/A)^{-1}}
$$
We recognize the quadratic form $b^\top(M/D)^{-1}b$. We know that $(M/D)$ is SPD, therefore $(M/D)^{-1}$ must also be SPD and its quadratic form $b^\top(M/D)^{-1}b > 0$. Therefore, the quantity inside the square root must be positive. Then, by chosing to add rather than subtract, we guarantee a solution with $d > 0$.
