I'm reading Madsen and Tornehave's book From Calculus to Cohomology.
Let $V$ be a vector space over $\mathbf R$. The vector space of all alternating 1-linear maps is denoted by $\text{Alt}^1(V)$ and can be given a basis $\pi_i:V\to\mathbf R$ of projection mappings, i.e. functions such that $\pi_i(v) = v_i$.
Now let $U\subseteq\mathbf R^n$ be open. The vector space of differential 1-forms, denoted $\Omega^1(U)$, consists of the smooth maps $\omega:U\to\text{Alt}^1(\mathbf R^n)$. If $x_i:U\to\mathbf R$ denotes the $i$th projection mapping in $\Omega^0(U)$, then $dx_i\in\Omega^1(U)$ is the constant map given by $x\mapsto\pi_i$.
Furthermore, $df = \sum\frac{\partial f}{\partial x_i}\pi_i = \sum\frac{\partial f}{\partial x_i}dx_i.$
My question is, are $\pi_i$ and $x_i$ the same map, even though they live in different places?
Sorry for writing a lot of words, but this also confused me when I was learning and I wanted to elaborate.
Differential forms are always confusing to introduce because the easiest example is to use a manifold ($U$ here) which is already a subset of $\Bbb R^n$. However, this makes it hard to distinguish coordinate charts from tangent spaces: for every $x\in U$, the tangent space $T_xU$ can be canonically identified with $\Bbb R^n$. So $U$ is a subset of $\Bbb R^n$ and every tangent space can be identified with $\Bbb R^n$.
This IS representative of the general picture, except generally $U$ is one small piece of a manifold $M$. Even if all of the tangent spaces $T_xU$ can be identified with $\Bbb R^n$, there is usually no way to simultaneously do this for every $x$ in $M$. (More precisely, there is a local trivialization $TM|_U\cong U\times \Bbb R^n$ but not between $TM$ and $M\times\Bbb R^n$.)
Because of this, it's best to think of each tangent space $T_xM$ as an abstract vector space which happens to be isomorphic to $\Bbb R^n$, but which doesn't show up with an obvious identification with $\Bbb R^n$ (in other words, $T_xM$ might not have an obvious choice of basis). In this spirit, the authors write $V$ (instead of $\Bbb R^n$) for an arbitrary tangent space and $\pi_i\colon V\to\Bbb R$ for a basis of the dual space (rather than $x_i$).
Long story short, $\pi_i$ (restricted to $U$) is indeed $x_i$ here, but I claim that this is not necessarily a useful perspective. Instead, I would imagine that, for every point $x\in U$, there is a tangent space $V$ which depends on $x$ and which varies smoothly for various $x$. The $\pi_i$ are local sections of the dual spaces $V^*$ and the $x_i$ are local coordinate functions on some subset of the manifold itself.
An example that might help is to consider $S^2$ embedded in $\Bbb R^3$. In this case the manifold locally looks like $\Bbb R^2$, but not globally. The tangent spaces can be locally identified with $\Bbb R^2$, but there is no way to do this globally (proving this fact is nontrivial, and is a consequence of the hairy ball theorem). Here you can see that there is some actual work to be done in identifying each tangent space with $\Bbb R^2$.