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.




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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 $\pi:TM\to M$. 


So I think the problem is that the splitting does not make sense if we do not have some metric at hand. And the vector field can not be defined like that.



A simple ODE on smooth manifolds
As you noticed, you need a connection $D$ on $TM$ in order to define the splitting in $T(TM)$ and the unique vector field $\xi$ with the property that $\xi$ is horizontal and satisfies $\pi\_\*(\xi\_X)=X$ for all $X\in TM$. Then the flow $\phi\_t$ of $\xi$ is the geodesic flow of the connection $D$. 


To see this, recall that from the definition of the covariant derivative, a vector field $Y$ along a curve $c$ in $M$ satisfies $D\_{\dot c}Y=0$ if and only if $Y$, wieved as a curve in $TM$, is horizontal (which means that $\dot Y$ is horizontal at each point wrt the above splitting).


Now, take any $X\in TM$ and denote by $\gamma\_t:=\phi\_t(X)$ and $c\_t:=\pi(\gamma\_t)$.
Since $\dot\gamma=\xi$, the curve $\gamma$ is the horizontal. Moreover, $\dot c$, wieved as a curve in $TM$, is just $\gamma$: 
$$\dot c\_t=\pi\_\*(\dot\gamma\_t)=\pi\_\*(\xi\_{\gamma\_t})=\gamma\_t.$$
By the above, $\dot c$ is $D$-parallel along $c$.


