Is there a "global" definition of differential $1$-forms? Let $M$ denote a smooth manifold. Then a covector at $p \in M$ is an element of the dual space of $T_p M$. We can organize covectors into a bundle over $M$, and then define a $1$-form on $M$ to be a section of this bundle.

Question. Is there a more direct approach to defining $1$-forms, like so:
A $1$-form on $M$ is a linear way of turning sections of $TM \rightarrow M$ into smooth functions $M \rightarrow \mathbb{R}$ satisfying some smoothness or "locality" conditions.
(I'm also interested in defining arbitrary $k$-forms in this way.)

 A: $\newcommand{\O}{\mathcal{O}}$
$\newcommand{\D}{\mathcal{D}}$
Let $\O_M=C^\infty(M)$ be the ring of smooth functions on $M$.
Let $\D_M$ be the set of $\Bbb R$-linear derivations of the ring
$\O_M$. We can identify the elements of $\D_M$ with the smooth tangent
fields on $M$. Then $\D_M$ is an $\O_M$-module, and we can identify
the smooth $1$-forms with the module $\text{Hom}_{\O_M}(\D_M,\O_M)$.
This point of view is developed systematically in the book
Smooth Manifolds and Observables by "Jet Nestruev" (Springer GTM 220).
A: You essentially said the same thing twice.  A covector is a way of turning a vector into a scalar.  Therefore a covector field (i.e., a differential 1-form on $M$) is a way of turning a vector field into a scalar field, i.e., a function on $M$.
A: As Mikhail Katz says, a $1$-form $\alpha$ is a machine that eats vector fields and spits out functions. Nothing more than that. As many others before me commented, you already wrote this in your post. If smoothness is the issue that bothers you, just add the requirement $$X\;\mathrm{is\;smooth}\Rightarrow\alpha(X)\mathrm{\;is\;smooth.}$$
