= Solution
Let $H$ be the horizontal lift of $iv$ supplied by the <canonical coframe of a surface unit tangent bundle>. The frame $(X,H,V)$ has commutators
$$
[V,X]=H,\qquad [V,H]=-X,\qquad [X,H]=KV.
$$
Since $F=X+\lambda V$, the <Lie bracket> product rule gives
$$
[F,V]=-H-(V\lambda)V,\qquad
[F,H]=-\lambda F+(K-H\lambda+\lambda^2)V.
$$
These formulas hold for every smooth $\lambda$, with no curvature or Anosov assumption.
Follow a tangent variation $\xi(t)=D\phi_t\xi(0)$ and express it in the moving frame as
$$
\xi(t)=a(t)F+y(t)H+z(t)V.
$$
Pulling this identity back by $D\phi_{-t}$ makes its left side constant. The derivative of the pulled-back frame vector $B$ is the pulled-back bracket $[F,B]$, so the coefficients satisfy
$$
\dot a=\lambda y,\qquad
\dot y=z,\qquad
\dot z=-(K-H\lambda+\lambda^2)y+(V\lambda)z.
$$
This derives the <linearized transverse equation for a surface thermostat>; its first equation in $a$ disappears on the quotient by $F$.
Near a vertical plane, $z\ne0$, and the smooth projective coordinate $r=y/z$ describes the plane as
$$
L=\mathbb RF+\mathbb R(V+rH).
$$
The <vertical Maslov cycle of a surface thermostat> is exactly $r=0$. Differentiating the quotient with the derived equations yields
$$
\dot r=1-(V\lambda)r+(K-H\lambda+\lambda^2)r^2.
$$
Consequently
$$
\boxed{F^*r=1\quad\text{on }\Lambda_V.}
$$
Since $r$ is a local defining function for the hypersurface, the nonzero normal derivative proves that \b[the lifted vector field is transverse to the Maslov cycle everywhere]. In particular, the sign follows from the forward derivative lift and the positive vertical-rotation convention, rather than from an unspecified projective orientation.
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