Understanding Schur complement through analogy
For a two-by-two matrix, you have
$$\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}e\\f\end{matrix}\right]
\quad\implies\quad
\left[\begin{matrix}a-bd^{-1}c&0\\0&d -ca^{-1}b\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}e-bd^{-1}f\\f-ca^{-1}e\end{matrix}\right].$$
This means that you can subtract a multiple of one row from the other row and obtain an equation of one variable, e.g.
$$(a-bd^{-1}c) x=e-bd^{-1}f,$$
from which point you can easily calculate the other variable by, e.g.
$y = d^{-1}(f - cx)$.
Schur complement is for two-by-two square block matrices, such as the block matrix $\mathbf M$ below. You can subtract a linear combination of one row from another row, and obtain a linear system
$$\left[\begin{matrix}\mathbf A&\mathbf B\\\mathbf C&\mathbf D\end{matrix}\right]\left[\begin{matrix}\mathbf x\\\mathbf y\end{matrix}\right] = \left[\begin{matrix}\mathbf e\\\mathbf f\end{matrix}\right]
\quad\implies\quad
\left[\begin{matrix}\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C&\mathbf 0\\\mathbf 0&\mathbf D-\mathbf C\mathbf A^{-1}\mathbf B\end{matrix}\right]\left[\begin{matrix}\mathbf x\\\mathbf y\end{matrix}\right] = \left[\begin{matrix}\mathbf e-\mathbf B\mathbf D^{-1}\mathbf f\\\mathbf f-\mathbf C\mathbf A^{-1}\mathbf e\end{matrix}\right].$$
Instead of solving the linear system involving the two-by-two block matrix ($\mathbf x$ and $\mathbf y$ at the same time), you could solve $\mathbf x$ and then $\mathbf y$.
The schur complement of $\mathbf A$ is written as $\mathbf M/\mathbf A=\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C$.
Subnet of an electric network
In terms of the equivalent electric network, one could write $\mathbf L\mathbf v = \mathbf i$. Let $V$ be all vertices, $B$ a subnet of vertices (target of study) and $\bar B$ all vertices outside the subnet. Then, one can rearrange the rows and columns to arrive at
$$\left[\begin{matrix}\mathbf L_{BB}&\mathbf L_{B\bar B}\\\mathbf L_{\bar BB}&\mathbf L_{\bar B\bar B}\end{matrix}\right]\left[\begin{matrix}\mathbf v_{B}\\\mathbf v_{\bar B}\end{matrix}\right] = \left[\begin{matrix}\mathbf i_{B}\\\mathbf i_{\bar B}\end{matrix}\right],$$
where the first block row gives $\mathbf L_{BB}\mathbf v_B + \mathbf L_{B\bar B}\mathbf v_{\bar B}=\mathbf i_B$. We can restrict our attention to the voltage in the subnet $\mathbf v_B$ by eliminating $\mathbf v_{\bar B}$ using Schur complement:
$$(\mathbf L_{BB}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf L_{\bar BB})\mathbf v_B=\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$$
The schur complement of $L$ over $\mathbf L_{BB}$ (resistance within subnet) is $\mathbf L/\mathbf L_{BB}$, which is $\mathbf L_{BB}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf L_{\bar BB}$, also called the response matrix. By applying the Schur complement $\mathbf L/\mathbf L_{BB}$ to the voltage of the subnet $\mathbf v_{B}$ would mean
$(\mathbf L/\mathbf L_{BB})\mathbf v_{B}$, which should generate $\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$.
Notation: $\mathbf L_{BB}=\mathbf L_\text{internal}$, $\mathbf L_{B\bar B}=\mathbf L_\text{mid}=\operatorname{Transpose}(\mathbf L_{\bar BB})$, $\mathbf L_{\bar B\bar B}=\mathbf L_\text{boundary}$.
Current through subnet boundary
... spits out the corresponding vector of current: how much electricity will flow through the boundary vertices.
This italic part must refer to $\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$. I'm not sure why this is the electricity flowing through the boundary vertices.
