I can't say for sure that it is the reason why the two notions are named the same, since they are not equivalent, but you have identified a relation between the two. In a way, the second notion is a special case of the first.
Let $(M,g)$ be a Riemannian manifold. As such, there exists a monomorphism of vector bundles
$$\flat : TM \to T^*M : v \mapsto v^{\flat} = v \, \lrcorner \, g \; .$$
Since both bundles have the same (finite) rank, it is also an epimorphism and so an isomorphism : there exists an inverse morphism
$$\sharp : TM \to T^*M : \lambda \mapsto \lambda^{\sharp} \; .$$
We can define a bundle metric $g^*$ on $T^*M$ as follows : for $\alpha, \beta \in T^*_mM$, let $g^*(\alpha, \beta) = g(\alpha^{\sharp}, \beta^{\sharp})$. This means that $\flat$ and $\sharp$ are "bundle isometries" : $(TM, g)$ is somewhat the same as $(T^*M, g^*)$. As such, the map $\flat$ sends the Liouville measure of and the geodesic flow on $(TM, g)$ to what we might consider the Liouville measure of and the geodesic flow on $(T^*M, g^*)$.
From a different point of view, $g^*$ defines a smooth "bundle quadratic form" $Q^*$ on $T^*M$ : given $\alpha \in T^*_mM$, let $Q^*(\alpha) = \frac{1}{2} \, g^*(\alpha, \alpha)$. This is a smooth real function of $T^*M$, in fact a "Hamiltonian" function if we equip $T^*M$ with its canonical symplectic form. There is an associated Hamiltonian vector field $X_{Q^*}$ whose flow preserves the level sets of $Q^*$, that is, the sphere-subbundles of $T^*M$ of different $g^*$-radii. Furthermore, this flow happens to be the same as the above "geodesic flow" on $(T^*M, g^*)$.
Since this flow preserves both the canonical volume form on $(T^*M, \omega_0)$ and the Liouville measure on $(T^*M, g^*)$, both volume forms have to be proportional to one another by a (nonvanishing) function $f$ which is constant along the flow. A priori, there is no reason for $f$ to equal 1 : after all, the canonical volume form knows nothing about the metric $g$ on $M$. However, in order to restrict the canonical volume form to a volume form on the level sets of $Q^*$, one needs to take the interior product with (for instance) a normal vector field, which necessitates a Riemannian metric on $T^*M$. I don't know if it is true, but may be there is a way to get the same measure on the level sets from both approaches.
If you take a look at the Wikipedia article on Liouville's theorem, you can read the following :
Although the equation is usually referred to as the "Liouville equation", Josiah Willard Gibbs was the first to recognize the importance of this equation as the fundamental equation of statistical mechanics. It is referred to as the Liouville equation because its derivation for non-canonical systems utilises an identity first derived by Liouville in 1838.
This suggests that the names "Liouville measure" might be only honorific too, since proofs of "the geodesic flow preserves the measure $\nu$" and "the Hamiltonian flow preserves the measure $\mu$" exist which use an identity of Liouville.
This is a good question. Here's how to do it: given coordinates $(x^1,\ldots, x^n)$ around (some particular point) $\gamma(t)$, where $\gamma\colon I \to M$, we have that $$\dot{\gamma}(t) = \sum_{i=1}^n \dot{x}^i(t)\frac{\partial}{\partial x^i}\bigg|_{\gamma(t)}$$for all $t \in I$. Then indeed $\nabla_{\dot{\gamma}(t)}\dot{\gamma}$ does not immediately makes sense, but we have the pull-back bundle $\gamma^*(TM) \to I$, and a connection $\gamma^*\nabla$. Then $\partial/\partial t$ is a vector field on the "base manifold" $I$, and $\gamma_\ast(\partial/\partial t) = \dot{\gamma}$, and this is what allows us to use the defining property of $\gamma^*\nabla$: $$\begin{align}\frac{D\gamma'}{{\rm d}t}(t) &= (\gamma^*\nabla)_{(\partial/\partial t)|_t}(\dot{\gamma}) = (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\sum_{j=1}^n \dot{x}^j \left(\frac{\partial}{\partial x^j}\circ \gamma\right)\right) \\ &= \sum_{j=1}^n \ddot{x}^j(t) \frac{\partial}{\partial x^j}\bigg|_{\gamma(t)}+ \sum_{j=1}^n \dot{x}^j(t) (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\frac{\partial}{\partial x^j}\circ \gamma\right) \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\dot{\gamma}(t)}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\sum_{i=1}^n \dot{x}^i(t) (\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{i,j=1}^n \dot{x}^i(t)\dot{x}^j(t) \nabla_{(\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{i,j,k=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} \\ &= \sum_{k=1}^n \left(\ddot{x}^k(t) + \sum_{i,j=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t)\right)\frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}.\end{align}$$
Best Answer
I've worked out my mistake - the vector field $X$ above is incorrect. The subtlety comes in raising and lowering the relevant indices.
Firstly we can all agree that the following vector field on the tangent bundle generates geodesic flow
$$Y = p^i \partial/\partial x^i - \Gamma^i_{jk} p^j p^k \partial/\partial p^i$$
where $(x^i, p^i)$ are coordinates on $TM$.
Now we want to use the tangent-cotangent isomorphism to determine a vector field on $T^*M$ which also gives geodesics after projection. Write out the ODEs for geodesic flow
$$\dot{x}^i = p^i $$ $$\dot{p}^i = -\Gamma^i_{jk}p^j p^k$$
Now define $p_i = g_{ij}p^j$ and calculate
\begin{align*}\dot{p}_i &= g_{ij,k}\dot{x}^k p^j + g_{ij}\dot{p}^j \\&= g_{ij,k}p^j p^k - g_{il}\Gamma^l_{jk}p^jp^k \\ &= p^j p^k(g_{ij,k} - g_{ij,k} + \frac{1}{2}g_{jk,i}) \end{align*}
Thus the correct vector field for geodesic flow on $T^* M$ is in fact
\begin{align*}X &= g^{ij}p_j \frac{\partial}{\partial x^i} + \frac{1}{2}\frac{\partial g_{jk}}{\partial x^i} p^j p^k \frac{\partial }{\partial p_i}\\ &= g^{ij}p_j \frac{\partial}{\partial x^i} - \frac{1}{2}\frac{\partial g^{jk}}{\partial x^i} p_j p_k \frac{\partial }{\partial p_i} \end{align*}
by raising and lowering indices, and using the derivative of the inverse metric. Now this agrees with $-dH$ in the question, after contracting with $\omega$.