How does one express a Lagrangian and Action in the language of forms? In Lipschitzs Classical Mechanics a Lagrangian is defined as:

$L(q,q',t)$ for some trajectory $q(t)$ of a particle

And the action is defined as:

$S:=\int^a_b L(q,q',t) dt$

How does one express this on the language of forms? Is the following possible?
Consider:

$L:TM \rightarrow \mathbb{R}$ 

so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a trajectory $q$ considered as curve, a 1-dimensional manifold:

$q:I \rightarrow J \subseteq M$. 

Then defining the action:

$S:=\int_J L$ 

we have $S=\int_{qI} L=\int_I q^*L=\int_I Lq_*$
Which we define as the Lagrangian $\int_I L(q,q',t) dt$ Introduced by Lipschitz.
Does this work?
 A: You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold.
Also to specifically address your question:

$L:TM \rightarrow \mathbb{R}$
so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate
over a trajectory $q$ considered as curve, a 1-dimensional manifold:

$q:I \rightarrow J \subseteq M$.


This part is not normally true, consider for example the free Lagrangian (evaluated at a point $p$ as a function $L_p : T_pM \to\mathbb{R}$) $L_p(v) = <v,v>_p$. This is definitely not linear in $v$, which is necessary for a 1-form. (Here < , > is some Riemannian metric on the manifold).
Also it may be that the set traced out by the curve is not a submanifold, this occurs in the case of a self-intersecting curve.
You do not need to have a 1-form to integrate over the curve. Since $L$ is a function $TM \to \mathbb{R}$, for a differentiable curve $\gamma : [a,b] \to M$ the composition: $L_{\gamma(t)}(\dot\gamma(t))$ is a function $[a,b] \to \mathbb{R}$ over which one can integrate.
Defining an action $S$ by $S[\gamma(t)]=\int_a^b L_{\gamma(t)}(\dot \gamma(t))\, dt$ one can recover the old equations of motion under the condition that $S$ is extremal for a physical path: By choosing any coordinate chart $\phi$ on a subset $U \subseteq M$, there is an induced set of coordinates on $TU \subseteq TM$ given by
$\tilde \phi (p,v) = (\{\phi(p)^i\},\{v_i\})$ where $v=\sum_i v_i\,  v(\phi(p)^i) \equiv \sum_i v_i \frac{\partial}{\partial\phi^i}$
On these coordinates $L$ gets the old form of a function $L(q,q')$ acting on two tuples of numbers and then e.o.m. $\frac{d}{dt}\frac{\partial L}{\partial q'}- \frac{\partial L}{\partial q} =0$ are recovered.
