For a Riemannian manifold $(M,g)$, the geodesic flow is $\phi_t:TM\to TM, (x,v)\mapsto (\gamma(t;x,v),\dot{\gamma}(t;x,v))$, where $\gamma(\cdot;x,v)$ is the geodesic started at $x$ with direction $v$. The related vector field of this flow can be formulated as $X(x,v)=(v,0)$.

My question can be viewed as the sham geodesic flow on manifolds without Riemannian metric. Namely, let consider the smooth manifold $TM$ and a vector field

$X:TM\to T(TM),(x,v)\mapsto(v,0)$.

By ODE we know this vector field can integrate to a (local) flow $\phi_t:TM\to TM$ (regardless any Riemannian metric). What is the meaning of this flow? Could we give an explicit formula of this flow?

In the simplese case $M=\mathbb{R}^d$, we know $\phi_t(x,v)=(x+t\cdot v,v)$ (indeed this is the geodesic flow with respect to the flat metric). I do not know if there is some similar form for flows on general manifolds.

Edit: Bill pointed out that the problem is not well formulated. I realized that I misunderstood the canonical splitting into horizontal and vertical parts:

$T(TM)=H\oplus V$ where the horizontal subspace $H$ is the kernel of the connection map $K : T(TM)\to TM$ and the vertical subspace $V = \mathrm{ker}(d\pi)$ is tangent to the fibers of $\pi:TM\to M$.

The vertical subspace $V$ does not involve any metric (hence I got the impression that $V\simeq TM$). At each point $\theta=(x,v)\in TM$, we have $T_\theta(TM)=H_\theta\oplus V_\theta\simeq T_xM\oplus T_xM$. But in general we may not have $V\simeq TM$ or $H\simeq TM$. Is this the point where I went wrong?