Geodesic vector field is well-defined Let $(M,g)$ be a Riemannian manifold. I just learnt that for a curve $x:I\to M$ to be a geodesic, the geodesic equation 
$$\ddot{x}^k+\dot{x}^i\dot{x}^j\Gamma^k_{ij}=0$$
is equivalent to the condition that the curve $(x,\dot{x})$ in $TM$ is an integral curve of the vector field
$$G=v^k\frac{\partial}{\partial x^k}-v^iv^j\Gamma^k_{ij}\frac{\partial}{\partial v^k}.$$
This is straightforward to verify, but I am unable to show that $G$ is a well-defined global vector field.

Question: How do you show that if $(\tilde{x},\tilde{v})$ are other coordinates for the tangent bundle $TM$, then
  $$\tilde{v}^k\frac{\partial}{\partial \tilde{x}^k}-\tilde{v}^i\tilde{v}^j\Gamma^k_{ij}\frac{\partial}{\partial \tilde{v}^k}=v^k\frac{\partial}{\partial x^k}-v^iv^j\Gamma^k_{ij}\frac{\partial}{\partial v^k}?\tag{1}$$
  i.e., in the language of physicists, I want to show that "$G$ transforms like a vector".

(I want to prove this without assuming existence and uniqueness of geodesics.) 
Attempt: Denote by $\tilde{G}$ and $G$ the left and right hand-side of $(1)$, respectively. Using the transformation
$$\tilde{\Gamma}^k_{ij}=\frac{\partial x^p}{\partial\tilde{x}^i}\frac{\partial x^q}{\partial\tilde{x}^j}\Gamma^m_{pq}\frac{\partial\tilde{x}^k}{\partial x^m}+\frac{\partial\tilde{x}^k}{\partial x^m}\frac{\partial^2x^m}{\partial\tilde{x}^i\partial\tilde{x}^j}$$
I get
$$\tilde{G}=G-v^rv^s\frac{\partial\tilde x^i}{\partial x^r}\frac{\partial\tilde x^j}{\partial x^s}\frac{\partial^2x^m}{\partial\tilde{x}^i\partial\tilde{x}^j}\frac{\partial}{\partial v^m}$$

But then, why does
  $$v^rv^s\frac{\partial\tilde x^i}{\partial x^r}\frac{\partial\tilde x^j}{\partial x^s}\frac{\partial^2x^m}{\partial\tilde{x}^i\partial\tilde{x}^j}=0?$$

 A: You seem to have used the wrong transformation rule for $\frac{\partial}{\partial x^i}$. In $TM$ it is not true that
$$\frac{\partial}{\partial\tilde x^i}=\frac{\partial x^j}{\partial\tilde x^i}\frac{\partial}{\partial x^j}\in\mathfrak{X}(TM)\tag{Wrong!}.$$
This is how $\frac{\partial}{\partial x^i}$ transforms as a vector field over $M$, but not as a vector field over $TM$. Indeed, in $TM$ we rather have
$$\begin{align}
\frac{\partial}{\partial\tilde x^i} &= \frac{\partial x^j}{\partial\tilde x^i}\frac{\partial}{\partial x^j}+\frac{\partial v^j}{\partial\tilde x^i}\frac{\partial}{\partial v^j} \\ 
&= \frac{\partial x^j}{\partial\tilde x^i}\frac{\partial}{\partial x^j}+\frac{\partial}{\partial\tilde x^i}\left(\frac{\partial x^j}{\partial\tilde x^k}\tilde{v}^k\right)\frac{\partial}{\partial v^j} \\
&= \frac{\partial x^j}{\partial\tilde x^i}\frac{\partial}{\partial x^j}+\frac{\partial^2 x^k}{\partial\tilde x^i\partial\tilde x^j}\frac{\partial \tilde x^j}{\partial x^l}v^l\frac{\partial}{\partial v^k}
\end{align}$$
Using this transformation, you will get an extra term that will cancel the one you thought should vanish.
