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.
Think back to what the Riemannian volume form is. It's a form that, when fed an oriented orthonormal frame, spits out 1. (Of course every manifold here is oriented.)
What are the oriented orthonormal frames on $\Sigma \subset M$, where $\Sigma$ is a (Riemannian) submanifold? Because it's a Riemannian submanifold, if $(x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p \Sigma$, it's also orthonormal in $T_p M$ - we're just missing one vector. By definition, $n$ is normal to all of these, and (modulo orientation conventions), $(n, x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p M$. Therefore $$i_nV_\Sigma(x_1, \dots, x_{d-1}) = V_M(n, x_1, \dots, x_{d-1}) = 1,$$ so that $i_nV_\Sigma$ does exactly as a volume form should.
You cannot refer to the induced volume form without at least having chosen a nonvanishing normal field. The Riemannian metric gives you a Riemannian structure on the normal bundle of $\Sigma$, which is what you need to define the induced volume form on $\Sigma$.
Best Answer
Let me first try to make sense of your question. First of all, the notion of a differential form is meaningless for general metric spaces $(X,d)$. However, one can still talk about Borel measures $\mu$ on the topological space $X$ (topologized using the metric $d$). The Borel condition still leaves too much freedom since we did not use the metric $d$ (only the topology). The most common condition these days is the one of a metric measure space which ties nicely $d$ and $\mu$ and allows one to do quite a bit of analysis on $(X,d,\mu)$ similarly to the analysis on the Euclidean $n$-space $E^n$. (The literature on this subject is quite substantial, just google "metric measure space".)
Definition. A triple $(X,d,\mu)$ (where $d$ is a metric on $X$ and $\mu$ is a Borel measure on $X$) is called a metric measure space if the measure $\mu$ is doubling with respect to $d$, i.e. there exists a constant $D<\infty$ such that for every $a\in X, r>0$, we have $$ \mu(B(a, 2r))\le D \mu(B(a,r)), $$ where $B(a,R)$ denotes the closed ball of radius $R$ centered at $a$.
Example. Every closed subset of $E^n$ (equipped with the restriction of the Euclidean metric) is a complete doubling metric space.
The basic existence result for doubling measures $\mu$, proven in "Every complete doubling metric space carries a doubling measure", by J. Luukainen, E. Saksman, Proc. Amer. Math. Soc. 126 (1998) p. 531–534, is the following:
Theorem. A complete metric space $(X,d)$ carries a doubling measure $\mu$ if and only if the metric $d$ is doubling, i.e. there exists a constant $C$ such that every ball of radius $2r$ in $X$ is covered by at most $C$ balls of radius $r$.
The proof of this theorem is not long but by no means trivial.