= Solution
The forcing is perpendicular to velocity, so
$$
\frac d{dt}|\dot\gamma|^2=2f\langle i\dot\gamma,\dot\gamma\rangle=0.
$$
Starting at unit speed therefore keeps the motion in $SM$. The horizontal part is the geodesic transport $X$, while the acceleration rotates the unit velocity at angular speed $f$. Thus the <magnetic flow> generator is $F=X+(f\circ\pi)V$.
For the <canonical coframe of a surface unit tangent bundle>, set $\mu=\alpha\wedge\theta\wedge\omega$. Its <interior product of a differential form> with the two relevant fields is
$$
\iota_X\mu=\theta\wedge\omega,\qquad
\iota_V\mu=\alpha\wedge\theta.
$$
The structure equations give
$$
d(\theta\wedge\omega)=d\theta\wedge\omega-\theta\wedge d\omega=0,
\qquad d(\alpha\wedge\theta)=0.
$$
The <Cartan formula for the Lie derivative> consequently gives $\mathcal L_X\mu=\mathcal L_V\mu=0$. For the variable vertical coefficient, use the full product rule:
$$
\mathcal L_{fV}\mu=d(f\iota_V\mu)
=df\wedge\alpha\wedge\theta+f\,d(\alpha\wedge\theta)=0.
$$
Indeed $f$ is pulled back from the base, so $Vf=0$ and $df$ has only $\alpha,\theta$ components. Hence
$$
\boxed{\mathcal L_F\mu=0,\qquad \phi_t^*\mu=\mu}.
$$
The <Liouville volume of a surface geodesic flow> is therefore preserved by this magnetic modification for every smooth $f$, without requiring the Anosov property.
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