= Solution
Use the derivative to transport the planes:
$$
\boxed{\phi_t^*((x,v),L)=\bigl(\phi_t(x,v),D\phi_t(x,v)L\bigr).}
$$
The derivative is an invertible linear map, so it sends a two-dimensional subspace to a two-dimensional subspace. The flow identity $\phi_t\circ\phi_s=\phi_{t+s}$ gives
$$
D\phi_tF(x,v)=F(\phi_t(x,v)).
$$
Hence a plane containing the generator is sent to another plane containing the generator; the displayed formula is a well-defined smooth lift to $\Lambda(SM)$. The <chain rule> gives $\phi_{t+s}^*=\phi_t^*\circ\phi_s^*$, and $\phi_0^*$ is the identity. Thus it is a genuine lifted <smooth flow>, with inverse $\phi_{-t}^*$, rather than merely a family of maps of individual fibres. Its generator is $F^*=\left.\partial_t\phi_t^*\right|_{t=0}$.
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