A vector field on the tangent bundle which is not equivalent to any second order ODE

Question

A second order differential equation on a manifold $M$ is a vector field $X$ on $TM$ which is not only a section of the vector bundle $T(T(M)) \to TM $ with the obvious structure, but also is a section of another bundle structure $(T(T(M)), TM, D\pi)$ where $\pi:TM \to M$ is the standard map and $D\pi$ is its differentiation.

What is an example of a real analytic vector field $ X$ on $TM$, the tangent bundle of a manifold $M$, such that its set of singularities is a discret set and is topological equivalent to NO second order vector field?

In particular, is there a polynomial vector field $X$ on $\mathbb{R}^2\approx T\mathbb{R}$ such that $X$ has a finite number of singular points(a generic case) but $X$ is not topological equivalent to any vector field in the following form? $$\begin{cases} x'=y \\ y'=g(x,y) \end{cases}$$

Answer 1

I guess you want the topological equivalence to preserve the bundle structure of $TM \longrightarrow M$ otherwise it becomes a bit arbitrary, right?

In this case a non-zero vertical vector field will not move the base points at all but still moves around in the fibers. If you have a second order differential equation then you necessarily move the points of the base or it is identically zero. Thus a nonzero vertical vector field should do the job: the existence is clear as you can take a vertical lift of a non-zero vector field on $M$.