Let me give a perspective from geometric algebra and calculus.
Geometric algebra, or clifford algebra, imposes a "geometric" product of vectors. If $a, b, c$ are vectors, then $a(b+c) = ab + ac$, and $(ab)c = a(bc) = abc$, so it's associative and distributive (and several vectors can be involved in a series of products).
Thus, the general objects of a geometric algebra are called multivectors. Components of a multivector are often separated into blades, where a blade of grade $k$ can be written as some geometric product of $k$ orthogonal vectors. As you might expect, when the base vector space of the GA is $\mathbb R^3$, then there are only 4 grades to consider: grade-0 (scalars), grade-1 (vectors), grade-2 (dubbed "bivectors"), and grade-3 ("trivectors," or in 3d, also called "pseudoscalars").
So you can see already that there is a relationship between the grades of multivectors and 0-forms, 1-forms, 2-forms, and 3-forms. A geometric algebra usually identifies forms with vectors through the usual inner product structure: if $w$ is a 1-form and $v$ is a 1-vector, then $w(v) \equiv w \cdot v$ is just the dot product. (Yes, perhaps the $w$ on the right isn't exactly the same kind of thing as the $w$ on the left, but I'm not aware what the standard notation might be for this concept.) While $k$-vectors and $k$-forms can still be said to transform differently under coordinate transformations, they're both still considered to be elements of the same geometric algebra.
So, having done away with one layer of distinction between $k$-vectors and $k$-forms, we can focus on how traditional vector calculus gets away with using only scalar and vector fields. As has been said, the key here is duality.
In differential forms parlance, this has to do with the Hodge star. In GA parlance, this has to do with the pseudoscalar, usually called $i$. In 3d, $ii = i^2 = -1$, so the notation is suggestive. Multiplication by $i$ changed the grade of what it acts upon. If $u = u_k$ is a $k$-vector, then $iu_k$ is a $3-k$ vector. So $i$ turns vectors to bivectors, scalars to pseudoscalars. This duality makes it possible to describe $2,3$-vectors in terms of their dual $0,1$-vectors--that is, wholly in terms of vectors and scalars.
Geometrically, you can picture a 2-vector field as a field of oriented planes through space. The dual vector field is just a vector field of normals to those planes. Similarly, a 3-vector field is a field of oriented volumes, but since all volumes are scalar multiples of one another, a scalar field contains all the same information (though it is not quite the same geometrically).