A short informal premise. As far as I understand, given a smooth map between manifolds $f: M \to N$ and a smooth vector field $X: M \to TM$, the pushforward $f_* X$ only defines in general a vector field along $f$, that is a map $f_* X: M \to TN$ such that $\pi_N \circ f_* X = f$. The pushforward $f_* X$ defines a unique vector field $Y: N \to TN$ if $f$ is a diffeomorphism or, more generally, if $f$ is surjective and $X$ and $Y$ are $f$-related, that is $f_*(m) X_m=Y_{f(m)}$ for each $m \in M$. The latter condition is fulfilled whenever $X$ is $f$-projectable, that is $f_*(m) X_m=Y_n$ for each $m \in f^{-1}(\{n\})$, where $f^{-1}$ denotes the inverse image.
These considerations can be applied to the case $f \equiv \pi_M : TM \to M$ and $X$ a vector field on $TM$. Since $\pi_M$ is surjective, if $X$ is projectable, then $\pi_{M*} X$ is a well-defined vector field on $M$ (the “projection” of $X$ onto the base manifold). In canonical local coordinates the most general vector field $X$ in $TM$ can be written as:
$$X=A^a(x,v) \frac{\partial}{\partial x^a} + B^a(x,v) \frac{\partial}{\partial v^a}$$
Since $(\pi_{M*} X)\psi :=X(\psi \circ \pi_{M})$ for $\psi :M \to \Bbb{R}$ (where the relation is to be intended pointwise), it seems to me that $X$ is $\pi_M$-projectable (and so $\pi_{M*} X$ is a vector field on $M$) iff $A^a$ only depends on $x$, that is for fields of the form
$$X=A^a(x) \frac{\partial}{\partial x^a} + B^a(x,v) \frac{\partial}{\partial v^a}$$
Am I correct?