Why generalize vector calculus with $k$-forms instead of $k$-vectors? The motivation usually given to differential forms is that they generalize vector calculus nicely. That's true, but there are also $k$-vectors, i.e., objects from $\Lambda^k(V)$ instead of $\Lambda^k(V^\ast)$. I particularly find their interpretation even easier to grasp and it sounds more natural as a generalization.
The problem that we usually use to say we need the generalization is a lack of cross product in higher dimensions, but $k$-vectors do possess a product exactly as $k$-forms do. Also, we could define everything: fields of $k$-vectors, derivatives of $k$-vectors and so on.
Why then usually the generalization is carried with $k$-forms instead of $k$-vectors?
 A: Because people want to be able to do calculus without the presence of a metric.
You know about $\nabla$ from vector calculus.  It's naturally written in terms of basis 1-forms.  If the basis one-forms are $e^1, e^2, \ldots$, associated with the coordinates $x^1, x^2, \ldots$, then $\nabla$ takes the form
$$\nabla = e^1 \frac{\partial}{\partial x^1} + e^2 \frac{\partial}{\partial x^2} + \ldots$$
For this reason, having $\nabla$ act upon a $k$-form field with a wedge product returns a $k+1$-form field.  Wedge products are non-metrical, so there's no problem.
Of course, you could consider a $k$-vector field and some analogue of the divergence:  if $A$ is a $k$-vector field, then we could dub $\nabla \cdot A$ as a $(k-1)$-vector field, and this is non-metrical because all we're really doing is having the basis 1-forms eat some vectors.  I can only speculate why this isn't done--perhaps because people want to be able to take gradients of scalar fields, and if you can only lower the grade of objects, you have to think a little harder about what a gradient is (it would really be a derivative of an $n$-vector field in an $n$-dimensional space).
Perhaps the biggest reason to avoid $k$-vector fields is integration, however.  This is something that traditional differential forms may gloss over.  What you're doing when you integrate a $k$-form is having it eat a $k$-vector representing the differential patch being integrated over.  That is, if $\omega$ is a $k$-form field and $m$ is a $k$-vector tangent to a $k$-dimensional manifold of integration $M$, then what's really going on is
$$\int_M \omega = \int_M \omega(m) dM$$
It might be clearer considering the 2d surface of integration case:
$$\int_M \omega(m) dM = \int_M \omega(m^{12} e_1 \wedge e_2 dx^1 dx^2 + m^{13} e_1 \wedge e_3 dx^1 dx^3 + \ldots)$$
A $k$-form eating a $k$-vector is, again, nonmetrical.  You couldn't do this without a metric if you tried to integrate a $k$-vector field, and you don't always have a metric.
That said, this "nonmetrical" approach often gets used in places where you do have a metric, and as a result, you end up going through hoops to avoid things you have at your disposal, which ends up quite silly.  It's perfectly valid to integrate a $k$-vector field on a Riemannian manifold, after all.
