You may find an elegant proof of this fact on **Paternain's book "Geodesic Flows"** (Birkhauser), in the very first pages. For convenience, I will reproduce the main parts of the argument here:

The most important step is to understand the geometry of $TTM$, the tangent bundle to the tangent bundle of $M$. Henceforth denote $\pi:TM\to M$ the footpoint projection. Note that, along the zero section, there is a canonical identification of $T_{(x,0_x)}TM=T_xM\oplus T_xM$, nevertheless this is not the case for arbitrary points in $TM$. Existence of a canonical horizontal complement to the vertical space $\ker d\pi$ is equivalent to having a connection.

## Connection map:
Fix a connection and consider the *connection map* $K:TTM\to TM$ defined as follows. Consider $\xi\in T_\theta TM$ and a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$. These give rise to a curve $\alpha=\pi\circ z$ (the projection of $z$ onto $M$)and a vector field $Z$ along $\alpha$, s.t. $z(t)=(\alpha(t),Z(t))$. Then $K$ is defined by $$K_\theta (\xi)=(\nabla_{\dot\alpha} Z)(0).$$

## Horizontal lift:
Now, define the horizontal lift $L_\theta:T_xM\to T_\theta TM$ as follows: given $v\in T_xM$ and $\beta:(-\epsilon,\epsilon)\to M$ a curve s.t. $\beta(0)=x$ and $\dot\beta(0)=v$, let $W(t)$ be the parallel transport of $v$ along $\beta$ and $\sigma:(-\epsilon,\epsilon)\to TM$ be the curve $\sigma(t)=(\beta(t),W(t))$. Then set $$L_\theta(v)=\dot\sigma(0)\in T_\theta TM.$$

Finally, we recall that in the above language, the geodesic vector field $G:TM\to TTM$ is clearly given by $G(\theta)=L_\theta(v)$.

## Symplectic structure of $TM$:

One can verify that, in the above notation, the canonical symplectic structure of $TM$ can be invariantly written as $$\omega_\theta(\xi,\eta)=g(d_\theta \pi (\xi),K_\theta(\eta))-g(K_\theta(\xi),d_\theta\pi(\eta)).$$

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>**Prop.** The geodesic field $G$ is the symplectic gradient of the Hamiltonian $H(x,v)=\tfrac12 g_x(v,v)$, i.e., for all $\theta \in TM$ and all $\xi \in T_\theta TM$, $$d_\theta H(\xi)=\omega_\theta (G(\theta),\xi).$$

**Pf.** With a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$, we have: $$d_\theta H(\xi)=\frac{d}{dt}H(z(t))\big|_{t=0}$$

$$=\tfrac12\frac{d}{dt}g_{\alpha(t)}(Z(t),Z(t))\big|_{t=0}$$

$$=g(K_\theta(\xi),v)$$

$$=g(d_\theta \pi(L_\theta (v)),K_\theta(\xi))$$

$$=g(d_\theta \pi(G(\theta)),K_\theta(\xi))$$

$$=\omega_\theta (G(\theta),\xi)\quad\square$$