Is there a coordinate-free proof of the hamiltonian character of the geodesic flow?

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: $$d_\theta H(\xi)=\frac{d}{dt}H(z(t))\big|_{t=0}$$

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

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

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

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

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

This is a very particular case of something more general: extremals for a non-degenerate Lagrangian correspond to solutions of the corresponding Hamiltonian system. There is no need to use connections at all. Check out Abraham and Marsden.

Here is an outline of a proof. It relies on the splitting of the tangent bundle of the tangent bundle $$TTM=VTM\oplus HTM$$ into the vertical part $VTM=ker d\pi,$ where $\pi\colon TM\to M,$ and into the Horizontal part, which is induced by the Levi-Civita connection as follows: Let $v\in T_pM$ and $\gamma$ be a curve centered at $p.$ Take the parallel section $v$ along $\gamma$ through $v$ wrt Levi Civita: Then, by definition, $v'(0)\in H_vTM\subset TvTM,$ and this construction gives a well-defined bundle complement of the vertical space in $TTM.$ Note that both $VTM$ and $HTM$ are canonically isomorphic to $\pi^*TM.$ This enables one to define a almost complex structure $J$ on TTM by switching in the appropriate way from $HTM$ to $VTM$ and vice versa. (I don't want to give a formula, as this would cause sign problems..). Moreover one gets an induced metric from M, which makes $HTM$ orthogonal to $VTM,$ denoted by $G.$

The first thing is to show that (lets call the symplectic form $\omega$) $$\omega(A,B)=\pm G(A,JB)$$ for all $A,B\in T_vTM.$ This again uses the definition of $HTM$ as well as Cartan formula. This can be done without coordinates.

The derivative of $K$ along the vertical space is just given as $$d_vK=G(\hat v,.),$$ where $\hat v\in V_vTM\subset T_vTM$ is the vertical vector corresponding to $\in T_pM.$ Therefore, the hamiltonian vector field $\xi$ of $K$ is given by $\pm J\hat v$ which is (after taking care with the sign) exactly the horizontal lift of $v\in T_pM$ to $H_vTM.$ An integral curve $\hat\gamma$ of $\xi$ is automatically a parallel section of $TM$ along its projection $\gamma=\pi\circ \hat\gamma,$ and as $\gamma'(t)=\hat\gamma(t),$ $\gamma$ must be a geodesic.