The reverse-mirror Seifert matrix is the negative original

Give the reflected surface $r(F)$ its transported orientation. Its boundary is $r(K)$ with the transported knot orientation. Therefore the Seifert surface for the specified reverse mirror is $F^{†}=-r(F)$, with the opposite surface orientation. Let $(u,v)$ be a positive tangent frame of $F$ and let $n$ be its positive normal, so $(n,u,v)$ is ambient-positive by [F1]. A positive tangent frame for $F^{†}$ is $(-dr(u),dr(v))$: reversing exactly one tangent reverses the surface orientation. Since $r$ reverses ambient orientation, the frame $(dr(n),-dr(u),dr(v))$ is ambient-positive. Consequently the positive normal of $F^{†}$ is $dr(n)$, and the positive push of the image basis curve $r(a_i)$ is precisely $r(a_i^{+})$. This computation includes both orientation reversals; reversing both tangent vectors instead would preserve the surface orientation.

Reflection negates linking. Indeed an oriented normal disk to an oriented curve $D$ is positive when its disk orientation follows the tangent of $D$ in the ambient orientation. The reflected disk has the opposite normal-disk orientation relative to $r(D)$, since $r$ reverses the ambient orientation. Thus $r$ takes the positive meridian class to minus the positive meridian class, and taking meridian coefficients in the complement gives $\mathrm{lk}(r(C),r(D))=-\mathrm{lk}(C,D)$. Using the actual positive pushes from step 1.1, the image-basis entries are therefore $\mathrm{lk}(r(a_i^{+}),r(a_j))=-\mathrm{lk}(a_i^{+},a_j)=-V_{ij}$. This proves the matrix is $-V$, including the empty matrix in genus zero.