= Solution
Let $\pi:SM\to M$ be the <unit tangent bundle> projection. Write $X$ for the <geodesic flow> generator and $V$ for the <vertical vector field of a surface unit tangent bundle>, positively rotating $v$ toward $iv$. The acceleration prescribed in the question changes only this velocity angle, so the <surface thermostat> generator is
$$
F=X+\lambda V.
$$
It never vanishes: $d\pi(F)=v\ne0$. Also $F$ and $V$ are linearly independent, since $d\pi(V)=0$.
The <vertical Maslov cycle of a surface thermostat> is
$$
\boxed{\Lambda_V=\{((x,v),L):V(x,v)\in L\}
=\{((x,v),\mathbb RF(x,v)\oplus\mathbb RV(x,v))\}.}
$$
There is exactly one such plane in each fibre, so $\Lambda_V$ is the image of a smooth section. A two-plane containing $F$ corresponds to a line in the rank-two quotient $T(SM)/\mathbb RF$; thus $\Lambda(SM)$ is a <projective bundle> with fibre $\mathbb{RP}^1$. Its total dimension is four and the cycle has dimension three, making it a smooth hypersurface. In this two-dimensional transverse space the <Maslov cycle> means the projective line meeting the distinguished vertical line, which here is just that distinguished line itself in each fibre.
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