Let us group the vertices as $U=\{u_1,u_2,\dots,u_n\}$ and $V=\{v_1,v_2,\dots, v_n\}$ where $f(u_i)=v_i$. Let $L_0$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, let $L _1 = \operatorname{diag}\left( \sum _{j=1}^n \mu(u _i, v _j)\right)$ and $L=L _0+L _1$,
also let $A$ be the symmetric matrix whose $a _{ij}$ is $-\mu(u _i,v _j)$. Clearly the Laplacian of $G$ is
$M=\left( \begin {array} {cc} L & A \\\
A & L \end {array} \right)$. Let $M^*$ stand for the matrix $M$ with deleted first row and column. We have
$$\kappa(G)=\det \left( \begin {array} {cc} L & A \\\
A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\\
D & (L+A)^ * \end {array} \right)$$
for some block matrices $B,C,D$ of size $n\times n,n\times (n-1),$ and $(n-1)\times n$ where this second matrix was obtained by adding the $i$th row of $M^{\*}$ to it's $n+i$th row for $1\le i\le n-1$ and then adding the first (or last) $n-1$ columns to the $n$th column. So $D$ is the matrix $(L+A)*$ together with a last column of zeros, making $(L+A)^{\*-1}D$ with integer entries.
Next we factor it using one of these identities
$$\kappa(G)=\det(L+A)^ * \det(B-C(L+A)^{*-1}D)$$
and observe that $L+A$ is the Laplacian of $G/f$ so $\det(L+A)^ *=\kappa(G/f)$, and since the second factor is an integer we get the desired divisibility.