(Below, by "vector space" I always mean "finite-dimensional real vector space.")
For $X$ any topological space there is a notion of a vector bundle over $X$ which formalizes the idea of a family of vector spaces parameterized by $X$. Vector bundles can be organized into a category similar to the category of vector spaces (which one recovers by taking $X$ to be a point); in particular one can define direct sums, duals, and tensor products of vector bundles.
An important notion here is that of a section of a vector bundle, which is roughly speaking a continuous choice of vector in each vector space in the family.
If $X$ is a smooth manifold (in particular if $X$ is an open subset of $\mathbb{R}^n$), then we can define a distinguished (smooth) bundle on $X$, the tangent bundle $T(X)$, coming from the tangent spaces at each point. The (smooth) sections of the tangent bundle are precisely vector fields on $X$. There is a dual bundle $T^{\ast}(X)$, the cotangent bundle, whose (smooth) sections are precisely differential forms on $X$.
So the tangent and cotangent bundles are in fact dual bundles, which means they have a dual pairing, and taking sections gives a dual pairing between vector fields and differential forms.
If in addition $X$ is compact, then we can make use of the Serre-Swan theorem, which identifies vector bundles over $X$ with finitely-generated projective modules over the ring $C^{\infty}(X)$ of smooth functions $X \to \mathbb{R}$. In this context I believe it's still true that the module of vector fields and the module of differential forms are dual, but I haven't worked out the details.
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.
Best Answer
This is a difficult question, one that I grappled with for a long time. Differential forms are simultaneously some of the easiest and most difficult concepts in basic mathematics.
Let me begin by saying how you think about Differential Forms is very dependent on how deep you want to go. If this is your first exposure I suggest that you look at this article by Tao. I don't think there is really anything I could say here in any short amount of time that would trump that.
On to your actual question. The relationship between the two objects, just like differential forms themselves, is a somewhat simultaneously sophisticated and simple idea. You have a bunch of questions all rolled into one in your post, and so I will try to address one. Namely, regardless of who you are talking to, a one-form is defined to be a section of the cotangent bundle of a manifold In less cryptic terms it is a mapping which associates to each $p\in M$ an element of $T_p^\ast M$. Thus, I'd like to explain how elements of $T_p^\ast M$ can be conflated quite literally with functionals on $T_p M$.
Let us first give a definition of tangent and cotangent space/bundle that will make the idea particularly clear.
Let $M$ be a smooth manifold of dimension $n$ and let $p\in M$ be a point. Let us then define an equivalence relation on the set $C^\infty(M)$ by declaring that $f\sim g$ if there exists a centered chart $(x,U)$ at $p$ such that $$\frac{\partial(f\circ x^{-1})}{\partial x_i}\mid_{v=0}=\frac{\partial (g\circ x^{-1})}{\partial x_i}\mid_{v=0}$$
for all $i=1,\cdots,n$. One can quickly verify that this definition is independent of chart choice. Moreover, one can prove that the set $T_p^\ast M:=C^\infty(M)/\sim$ inherits a vector-space structure by merely declaring that $\alpha[f]+\beta[g]:=[\alpha f+\beta g]$ where $[h]$ denotes the equivalence class of a function $h\in C^\infty(M)$ under $\sim$. This resulting vector space is called the cotangent space of $M$ at $p$.
Similarly, let us define a space which, in a sense, is just the cotangent space but with the arrows reversed. Namely, let us denote by $\mathcal{F}_p$ the set of all smooth functions $f:U\to M$ where $U$ is a neighborhood of $0$ in $\mathbb{R}$ and $f(0)=p$. Let us then define $\mathcal{G}_p$ to be the set of germs $\mathcal{F}_p/\simeq$ where $f\simeq g$ if $f$ and $g$ agree on some neighborhood of $0$ (we're dealing with derivatives which only care about the function locally, so we don't care about a function except with respect to how it behaves locally at $0$--so we identify two functions that are indistinguishable near $0$). We shall denote the germ of a function $f\in\mathcal{F}_p$ by $\{f\}$. Lastly, let's define $T_p M:\mathcal{G}_p/\approx$ where $\{f\}=\{g\}$ if there exists a chart $(x,U)$ centered at $p$ such that
$$(x\circ f)'(0)=(x\circ g)'(0)$$
for all $i=1,\cdots,n$ (note that since $x\circ f$ and $x\circ g$ are functions from a subset of $\mathbb{R}$ to a subset of $\mathbb{R}^n$ this just means that the derivative of each of the coordinate functions of $x\circ f$ and $x\circ g$ at $0$ agree). Let us denote an equivalence class of $f\in\mathcal{F}_p$ (really of $\{f\}\in\mathcal{G}_p$) by $\underline{f}$. We see then that, at least as a set, $T_p M$ really is just like $T_p^\ast M$ but with all of the arrows reversed. We can define a vector space operation on $T_p M$ by saying that $\alpha \underline{f}+\beta\underline{g}=x^{-1}(x\circ f\circ m_\alpha+x\circ g\circ m_\beta)$ where $m_\gamma$ is multiplication by $\gamma$. I leave it to you that this is really a vector space operation--it's just basically so that the derivatives work out to add values and multiply values correctly.
Now, while the above definitions make look non-standard to you (for example you may have had $T_p M$ defined for you in terms of derivations at $p$ of $C^\infty(M)$) you can check that these vector spaces are canonically isomorphic to whichever definition you used. So, why did I use them? Because they highlight a very natural pairing $\langle -,-\rangle:T_p M\times T_p^\ast M\to \mathbb{R}$. Namely,
$$\langle \underline{f},[g]\rangle=(g\circ f)'(0)$$
I leave it to you that not only is $\langle-,-\rangle$ well-defined (independent of representative from the equivalence classes) but that it is actually bilinear and non-degenerate. Note that by how we have defined $T_p M$ and $T_p^\ast M$ this pairing is absolutely natural.
So what? What does this extremely natural pairing give us? Well, up until this point we have had one way to think about elements of $T_p^\ast M$--as just elements of $T_p^\ast M$ (equivalence classes of functions)! But, this bilinear pairing allows us to define a very natural map $T_p^\ast M\to (T_p M)^\ast$ (where this right-hand side is the vector dual space of $T_p M$)by taking $[g]$ to $\langle -,[g]\rangle$. The bilinearity and non-degenerateness are precisely what we need to show that this mapping is a well-defined linear isomorphism. Because this isomorphism is so natural (coming from such a natural pairing) this allows us to blur the lines between $T_p^\ast M$ and $(T_p M)^\ast$.
This ability to basically treat $T_p^\ast M$ as $(T_p M)^\ast$ and vice-versa is why people rarely make a fuss when thinking about one-forms as being either associations of points to cotangent vectors or the association of points to linear functionals eating in tangent vectors.
The highfalutin way of phrasing the above is that a one-form is a section of the cotangent bundle $T^\ast M$. But, using the above ideas you can show that the cotangent bundle is isomorphic to the dual bundle $(TM)^\ast$ of the tangent bundle $TM$. This is where all of the naturality comes in--we needed the isomorphism $T_p^\ast M\cong (T_p M)^\ast$ (which is clear by dimension counting) to not rely on choosing a basis so that the fiberwise isomorphism pieces together smoothly.
As I pointed out above Terry Tao is the right man to get an intuition for what one-forms are. How to practically deal with them, which definition to practically take is more difficult. The simple answer is that which situation you are in shall dictate which idea of one-forms is most natural. You should invest time in understanding, and being comfortable with all of theses different ways of thinking about one-forms and how to move seamlessly between them. It will not only make your life easier now, insomuch as that you will erase any question of right definition (they are all right!) but will be absolutely crucial once you start using one-forms in the future.