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.
In linear algebra it is important to understand the distinction between a linear transformation $T:V\to W$ and a matrix $A$ which represents $T$. Different choices of bases for $V$ and $W$ will give different matrices. The transformation $T$ is "coordinate-free" but the matrix $A$ is not.
Another example is an inner product on a vector space. It exists independently of any choice of basis ("coordinate-free") but if we choose a basis we can represent the inner product using a matrix (which would not be coordinate-free).
Differential forms are coordinate-free in the same way. They exist independently of any choice of basis but if we choose we can express them using a basis.
Why is a coordinate-free formulation superior? Sometimes equations look simpler. For example, compare the coordinate-free geodesic equation to the coordinate-dependent geodesic equation:
$$\nabla_{\gamma'}\gamma'=0$$
vs.
$$\frac{d^2\gamma^k}{dt^2}+\Gamma_{ij}^k\frac{d\gamma^i}{dt}\frac{d\gamma^j}{dt}=0.$$
Or compare the coordinate-free equation for closedness of a differential form ($d\omega=0$) to what it is written out in coordinates.
Also, from a philosophical perspective, coordinates are rather arbitrary (nature doesn't come with axes), so it is more pleasing if we can write something in a coordinate-free way.
Best Answer
The thing about $\mathbb{R}^n$ is that it "naturally" comes with a choice of coordinates since $\mathbb{R}^n = \overbrace{\mathbb{R}\times \cdots \times \mathbb{R}}^n$ this means that elements of $\mathbb{R}^n$ are of the form $\mathbf{x}=(x_1,...,x_n)$. We also have maps
\begin{align*} x_i &=\text{pr}_i : \mathbb{R}^n \rightarrow \mathbb{R}\\ &\mathbf{x} \mapsto x_i \end{align*}
These maps are also natural and they do not need the introduction of more objects. And the basis one forms are the differentials of these maps, so technically we haven't chosen coordinates we've only used the definition of $\mathbb{R}^n$. We can use different coordinates for $\mathbb{R}^n$ though, say for the case $n=2$ we can use polar coordinates (though there is a slight subtlety at the origin). In this case we have the basis forms $dr$ and $d\varphi$, and if we use the rule of how differential forms change under change of coordinate $ (r,\varphi)\mapsto (x,y)$ we find
$$dx=\cos\varphi dr - r \sin \varphi d\varphi \hspace{15mm} dy = \sin\varphi dr + r \cos \varphi d\varphi$$
From here we notice that $dx\wedge dy = r dr \wedge d\varphi$ (technically I should have written pullbacks but that's beside the point). We see that we obtain the element of area in polar, so the area of a region given in polar coordinates is the same as it would have been if I used cartesian. This is the reason why coordinate independence is useful because I can define an object $\omega = dx\wedge dy$ in cartesian, and write an expression like
$$\int_D \omega =\text{Area}(D)$$
And that expression does not depend on whether I use polar or cartesian coordinates.
A lot of the definitions can be made into coordinate independent ones, some of them are not as popular as the coordinate versions though. For example you have probably seen the definition for the exterior derivative of a one form $\alpha= \alpha_j dx^j$ in local coordinates as
$$d\alpha = d\alpha_j \wedge dx^j = \partial_i \alpha_j dx^i \wedge dx^j$$
There is however a coordinate independent definition for the exterior derivative, the one for one forms becomes
$$d\alpha(X,Y) = X(\alpha(Y))- Y(\alpha(X))- \alpha(\left[X,Y\right])$$
Hope this helps.