At the end of page 280 of Lee's Introduction to Smooth Manifold the author states
Although the partial derivatives of a smooth function cannot be interpreted in a coordinate-independent way as the components of a vector field, it turns out that they can be interpreted as the components of a covector field. This is the most important application of covector fields.
What does the above passage mean? It would help if someone could rephrase what the author means more rigorously.
In the next passages, the author shows that, given smooth coordinates $(x^i)$, we may write $$\text{d}f = \frac{\partial f}{\partial x^i}\text{d}x^i.$$
Is this what the author meant in the quoted passage? If so, in what sense is the above equality 'coordinate-independent'?
Given a coordinate chart, $(U,x)$, I hope you know that there is the corresponding chart-induced vector fields on $U$, namely $\left\{\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right\}$. For each $p\in U$, these vector fields evaluated at $p$ give a basis for $T_pM$. Next, for any covector field $\omega$ on $U$, it has components relative to the above basis simply by evaluating the covector field on each of the basis vector fields, namely $\omega_{(x),i}:=\omega\left(\frac{\partial}{\partial x^i}\right)$ (perform a pointwise evaluation on $U$; this gives us a function $\omega_{(x),i}:U\to\Bbb{R}$). From these components, we can recover the full covector field simply as $\omega=\omega_{(x),i}\,dx^i$. If this last step is not clear then you should revisit the situation in a single vector space $V$. Take a basis $\{e_1,\dots, e_n\}$ for $V$, construct the dual basis $\{\epsilon^1,\dots, \epsilon^n\}$, and prove that for each $v\in V$, we have $v=\sum_{i=1}^n\epsilon^i(v)e_i$ and for each $\lambda\in V^*$, we have $\lambda=\sum_{i=1}^n\lambda(e_i)\epsilon^i$; the $\epsilon^i(v)$ and $\lambda(e_i)$ are typically just written $v^i,\lambda_i$ respectively.
Now, you can play this game with $\omega=df$. By unwinding some definitions, you will see that the component $(df)_{(x),i}$ is exactly the $i^{th}$ coordinate partial derivative $\frac{\partial f}{\partial x^i}$, and by applying the above linear algebra fact, we get out beloved $df=\frac{\partial f}{\partial x^i}\,dx^i$.
Regarding coordinate-independence, it simply means that the defintion of $df$ doesn’t depend on coordinates. Using coordinates, it can be built out of things which do depend on coordinates (namely $\frac{\partial f}{\partial x^i}$ and $dx^i$ separately) but their specific linear combination is $df$ and that doesn’t depend on coordinates.