What exactly is a 0-form? From what I understand, a k-form in the real numbers is essentially a mapping $\mathbb{R^k} \rightarrow \mathbb{R}$, but I can't seem to find a corresponding definition for a "0-form". Wikipedia seems to call it a smooth function "on" some set $U \subset \mathbb{R^n}$, but this is confusing me. Does this mean a mapping $\quad f: U \rightarrow U$, or $f: \mathbb{R} \rightarrow U$, or something else?
 A: A zero form is a smooth function defined on a manifold. Here smoothness means that the map $U\rightarrow \mathbb{R}$ is smooth by considering the smoothness of chart maps $\phi:\mathbb{R}^{n}\rightarrow U\rightarrow \mathbb{R}$. In general an $n$-form is a section of the anti-symmetric cotangent tensor bundle $\wedge^{n}(TM^{*})$. 
A: The notion of a '0-form' is just a convenient device. Basically k-form means for a fixed point $x$ you pick k-vectors starting at this point $x$ and pointing elsewhere. That is, you pick k number of vectors from the tangent space of some point $x$
$$f(x)\{(x;v_1), ... , \underbrace{(x; v_k)}_{\text{we pick k vectors starting at point x and ending at point $v_k$ of some $\mathcal{R}^n$}}\}$$, which is a real-valued function. Zero means we pick no vectors from the tangent space, we assign to x some scalar (=real) value.
A: In general, a $k$-form on a vector space $\mathbb{V}$ over a field $\mathbb{F}$ is a multilinear, alternating map that takes $k$ vectors and returns a scalar (an element of $\Bbb F$). So a $0$-form is a map that takes no vectors at all and returns a scalar: We can concretely think of a $0$-form as a map $f : \Bbb F \to \Bbb F$, and for many purposes we may as well just identify this map with the scalar $f(1)$ itself.
On an open set $U \subset \mathbb{R}^n$, a smooth $k$-form is a smoothly varying choice of $k$-form on each tangent space $T_p U$, and so we may identify a smooth $0$-form $f$ with the smooth function given by mapping each $p$ to the scalar corresponding to the $0$-form $f_p$.
(To make clear the distinction between these two kinds of objects, we occasionally call the latter a $k$-form field on $U$.)
A: I'm a little late to the party, but I hope I can make this a little more understandable. 
A differential $k$-form on $\mathbb{R}$ is more like a mapping from $\mathbb{R}\times\mathbb{R}^{k}$ to $\mathbb{R}$. Thus, a $0$-form on $\mathbb{R}$ is a smooth function from $\mathbb{R}\cong\mathbb{R}\times\mathbb{R}^0$ to $\mathbb{R}$. This is actually a result of convention, where, in the general case of "smooth manifolds," we shrewdly choose $0$-forms to be smooth functions.
Also, I would be careful talking about $k$-forms on $\mathbb{R}$. While they exist, things get a little bit silly for any $k>1$.
