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In my course about differential forms, we define 1-forms as follows:

If $(e_1,..,e_m)$ is the standard basis of $\mathbb{R}^m$ and $\sigma$ a chart around $p\in M$ on the m-dimensional manifold $M$, then $d\sigma_p(e_1),..,d\sigma_p(e_m)$ is the standard basis of the tangent space $T_pM$, and the dual basis for the cotangent space is $dx_1(p),..,dx_m(p)$ where $x_i$ is the coordinate function $x_i(p)=x_i(\sigma(u))=u_i$.

Hence, having a basis for the dual space, we may write any differential in this basis. For example if $M=\mathbb{R}^2$ and $f$ smooth function: $\mathbb{R}^2\rightarrow \mathbb{R}$, then we write $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

Now comes the question:

considering polar coordiantes $(r,\phi)$, we want to express $dx,dy$ with $r,\phi,dr,d\phi$. So I would like to justify the fact that $dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \phi}d\phi$. Why is that true? Why is $(dr,d\phi)$ a basis of the cotangent space and why are the coefficients in that basis the partial derivatives with respect to polar coordinates?

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You already wrote $df$ for some scalar field $f$ in terms of Cartesian coordinates. The expression in terms of polar coordinates (or any other coordinates) is actually quite similar.

$$df = \left( dx \frac{\partial}{\partial x} + dy \frac{\partial}{\partial y} \right) f = \left( dr \frac{\partial}{\partial r} + d\phi \frac{\partial}{\partial \phi} \right) f$$

Put $f = x(r, \theta)$ in, and you get the result.

Personally, I think the $dx, dy, \ldots$ notation for basis covectors is hideous, and it may be part of what's confusing you here. Try denoting the basis covectors with something else (I've seen thetas used, i.e. $\theta^x, \theta^y, \theta^r, \theta^\phi$, or even $e$'s).

Geometrically, you should be able to realize that basis covectors tell us about normal directions to surfaces of constant coordinates. $dx$ is normal to a surface of constant $x$; $dr$ is normal to a surface of constant $r$, and $d\phi$ is normal to a surface of constant $\phi$. Imagining the latter two, it should be clear that $dr, d\phi$ can and do form a basis at any given point other than the origin.

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