For 1-forms, you can get some intuition from Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:
Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.
Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form hypersurfaces in $M$. In this case, it turns out that one can fit the planes together if and only if $d\alpha\equiv0$. (In the general case, where instead of $\alpha$ we have an algebraic ideal of 1-forms $\mathcal I$, this is $d\mathcal I\equiv 0\mod\mathcal I$).
Here's some more discussion: if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const. Similarly, if $d\alpha=0$ then it's clear that we can find such $f$ locally (not globally if $\alpha$ isn't exact).
Hence $d\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.
(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners).
Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $d\alpha = dx dy$:
