Understanding Differential Forms as elements of a dual space So I have studied quite a bit of differential geometry and am comfortable with differential forms and related notions. Of late however, I have been spending time seeing if I actually understand the objects/concepts that I am using all the time. This has led me to investigate whether I understand what a differential form is. For our purposes here, let us suppose that everything is smooth. 
Originally, I had been satisfied with considering a differential form as a smooth section of the cotangent bundle, or as a basis vector for the cotangent space. This means however that for each point $p \in M$, where $M$ is a smooth manifold, a differential form $$\omega = \sum_{i_1 \cdots i_n =1}^n f_{i_1 \cdots i_n} dx^{i_1} \wedge \cdots \wedge  dx^{i_n}$$ assigns to $p$ a corresponding linear functional, call it $\omega_p$. 
I want to make sure that I'm not just symbol pushing and actually understand this properly. Take for example, the 1-form $\omega = f(x) dx$, where $f$ is smooth. Then $\omega$ maps $p$ to the linear functional $\omega_p = f(p)dx$? I don't see how $f(p)dx$ is a linear functional. 
 A: I think you may have forgotten what the symbol $dx$ means.  Here $x$ is one of the coordinate functions in a chart at $p$, so it is a smooth function from some neighborhood of $p$ to $\mathbb{R}$.  Then $dx$ is by definition the total derivative of $x$: that is, it is the functional on the tangent space at $p$ which takes a tangent vector $v$ at $p$ and outputs the directional derivative of the coordinate function $x$ with respect to $v$.  Explicitly, if your local coordinates are $(x_1,\dots,x_n)$ with $x=x_i$ for some $i$ and you think of $v$ as a vector $(a_1,\dots,a_n)\in\mathbb{R}^n$ using these local coordinates, then $dx(v)$ will just be $a_i$.  The functional $f(p)dx$ is then just this functional $dx$ multiplied by the scalar $f(p)\in\mathbb{R}$.
A: The linear functional $\omega_p=f(p)\,dx$ eats a vector like $\partial_x$. We have $dx(\partial_x)=1.$ If $v=g(x)\partial_x$, then $\omega_p(v)=f(p)g(p)\,dx(\partial_x)=f(p)g(p).$ 
The general action of a 1-form $df$ on a vector $v$ is given by the cute equation
$$df(v)=v(f).$$
