Is there an easy way to prove that this matrix is total unimodular ? $$ \begin{bmatrix} 1 & F_1 & 0\\ 1 & 0 & F^T_1 \\ 0 & F_2 \end{bmatrix} $$
$1$ is the identity matrix, and $F_1, F_2$ are network matrices (so per row they have one +1 and one -1 entry or only one +1 entry) [$F^T_1$ is the transposed matrix, $F_2$ is as big as $[0 \; F^T_1]$ ].
I already proved that the matrix without the row $[1 \; 0 \; F^T_1]$ is total unimodular.
The matrix is not totally unimodular, in general. A simple counterexample is if $F_1=[1\,-1]$. Then your matrix contains as a submatrix $[1\, F_1^T]$. This is $$\left[\begin{array}{rr} 1 & 1\\ 1 & -1 \end{array}\right]$$ which has determinant 2.