I have just started studying differential forms and I understand that 0-forms are smooth real valued functions defined on $\mathbb{R}^n$, 1-forms are linear combinations of $\mathrm{d}x_i$ with smooth functions from $\mathbb{R}^n$ to $\mathbb{R}$ being coefficients, 2-forms are linear combinations of $\mathrm{d}x_i$∧$\mathrm{d}x_j$ with coefficients being smooth functions from $\mathbb{R}^n$ to $\mathbb{R}$ and so on. I understand that $\mathrm{d}x_i$, is the function that takes as input a vector in $\mathbb{R}^n$ and returns it's $\mathrm{i}^{th}$ component. So is $\mathrm{d}x_i$ a 0-form or a 1-form? Because by the definition of $\mathrm{d}x_i$, it is a smooth function from $\mathbb{R}^n$ to $\mathbb{R}$ so it should be a 0-form. At the same time it can also be considered as a 1-form. And if the answer to the above is that it is both a 0-form and a 1-form, then while taking pullbacks, why do we define the pullback of 0-forms separately and the pullback of $\mathrm{d}x_i$ differently? Is it because the pullback of $\mathrm{d}x_i$ as 0-form should be one thing and its pullback as a 1-form should be another? That seems a little strange to me.
I am looking at Ted Shifrin's lectures if that helps.
Note that $dx_i$ does not take a vector $p\in \Bbb R^n$ and spits out its $i$th component. That would be the scalar function $x_i$. Instead, $dx_i$ takes in a vector field $X$, and for each point / vector $p\in \Bbb R^n$ it takes the vector $X_p$ of the field at $p$ and spits out the $i$'th component. Thus $dx_i$ makes a vector field into a scalar function.
In general, a $k$-form takes in $k$ vector fields and spits out a scalar funciton (there are other requirements too, like antisymmetry and smoothness and such). This makes $dx_i$ into a $1$-form.
Finally, $x_i$, which we recall is the scalar function given by the $i$'th component of the position vector, is a $0$-form, and $d$ always increases the degree of a form by exactly $1$. More generally, for a scalar function $f$, $df$ is the $1$-form that takes in a vector field $X$, and at each point $p\in \Bbb R^n$ gives the directional derivative of $f$ at $p$ along the vector $X_p$.