I stumbled across this paper providing an intuition of differential forms. On page $4$ the paper reads
"$df$ is represented as the contour lines of $f$ with the contours corresponding to equally spaced 'heights'"
Why is this the case?
An explanation in the comments of this post goes as follows:
"Think about what this means in a coordinate. Then $d$ (in two dimensions) acts on $f$ by $df=∂xf dx+∂yf dy$. This is nothing more than $∇f⋅(dx,dy)$, so at any given point, you have something which looks roughly like the picture in the OP, but parallel to the gradient of $f$. You get the contour lines because this $1$−form is actually the derivative of something, so they can be integrated, i.e. "smoothed out" into a continuous contour line."
to what I answered in the comments:
"I understand that $df=∇f⋅(dx,dy)$, but how does that imply that 'at any given point you have somthing which looks roughly like the picture in the OP, but parallel to the gradient of f'?"
The way to think about is, to increase "x" we have to move across horizontal lines. Similarly to increase "f", we have to move along contour lines. In this question, you see that contour lines of the function $g(x,y)=x$ is just horizontal lines the $x-y$ plane . Similarly to increase $f$, we need to cut across contour lines of $f$.
If we are at a point in the domain, then the contour lines which we cut of $f$ is propotional to the projection of our "step" along the direction perpendicular to our contour.
Shortly put, $df(v) = \nabla_v f$