Let $M$ be some manifold, $x$ a chart on $U\subset M$ and $\omega$ a $p$-form such that \begin{equation} \omega=\omega_I\wedge dx{}^I:=\sum_{1\leq i_1<\ldots<i_p\leq n}\omega_{i_1,\ldots,i_p}dx^{i_1}\wedge\cdots\wedge dx^{i_p} \end{equation} on $U$. From The Geometry of Physics, page $141$:
In any manifold the operation of exterior differentiation [...] can be written symbolically as $d = dx^j\wedge\partial/\partial x^j$; the operator $\partial/\partial x^j$ acts only on the coefficients.
Here is my attempt of a proof: We have \begin{align} &d\omega=d(\omega_Idx^I)=d(\omega_I\wedge dx^I)=d\omega_I\wedge dx^I+\omega_I\wedge ddx^I\\ &=d\omega_I\wedge dx^I+0=d\omega_I\wedge dx^I=d\omega_I\wedge dx^I=\partial_j\omega_Idx{}^j\wedge dx^I\\ &=dx{}^j\wedge \partial_j\omega_Idx^I=dx^j\wedge\partial_j\omega \end{align} where we have defined $\partial_j(\omega_Idx^I):=(\partial_j\omega_I)dx^I$.
Is this correct? In particular, is it allowed to write $\omega_I dx^I=\omega_I\wedge dx^I$?
Comment: The definition of $\partial _i\omega$ might seem a bit weird, but I noticed that it also appears in other parts of the book:
From the overview before the first chapter:
The electric field intensity is a $1$-form $\mathscr E$ and the magnetic field intensity is a $2$-form $\mathscr B$ and Faraday’s law says \begin{equation} d\mathscr E = −\frac{\partial \mathscr B}{\partial t} \end{equation} where $\partial \mathscr B/\partial t$ means take the time derivative of the components of $\mathscr B$.
From page $119$:
We define the time derivative of an exterior form by differentiating each component\begin{equation} \frac{\partial}{\partial t}\omega:=\frac{\partial\omega_I}{\partial t}dx^I \end{equation}
Thus I thought that maybe we can define $X\omega$ - where $X$ is a vector field and $\omega$ is a $k$-form - to be the $k$-form $X\omega:=X(\omega_I)dx^I$. If we can prove that the definition doesn't depend on the coordinate system, this would be well defined...
Your $X\omega$ is not well defined. It's coordinate dependent. Consider the coordinate systems $x$ and $y$ in which $y = f(x)$ for some non-constant increasing and nonvanishing function (the manifold is $\mathbb{R}$ here). Consider the 1 form $\omega := dy = f'(x) dx$. Let $X = \frac{\partial}{\partial y} = \frac{1}{f'(x)} \frac{\partial}{\partial x}$.
Then $X(dy) = \frac{\partial}{\partial y} (1) dy = 0$
And $X(dy) = \frac{1}{f'(x)} \frac{\partial}{\partial x} f'(x) dx = \frac{f''(x)}{f'(x)} dx \neq 0 $.
There is indeed an invariant formula for the exterior derivative. If $\omega$ was a 2-form: then
$$d\omega (X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$$ for any vector fields $X,Y$. The general invariant formula looks a little more complicated. Check it out on Wikipedia:
https://en.wikipedia.org/wiki/Exterior_derivative#In_terms_of_invariant_formula.
EDIT: Your formula for $d\omega$ is correct. In local coordinates, $$d\omega = dx^i \wedge \frac{\partial}{\partial x^i} \omega_I dx^I$$ which is indeed coordinate independent. That is how many authors define the exterior derivative actually. They define it locally in that way and show that this expression is coordinate independent, which shows that $d\omega$ is well defined globally. However, $\frac{\partial}{\partial x^i} \omega_I dx^I$ is not coordinate independent and so is not a well defined global 1-form as I show above.