Questions about differential forms Let $\omega$ be a $k$-form on $M \subseteq \mathbb R^n$ a $d$-dimensional submanifold of $\mathbb R^n$ and $k,d \le n$.
1) Can I only integrate $\omega$ over $M$ if k = d? Why?
2) Let $\omega$ be a $1$-form and $\alpha: \mathbb R  \to \mathbb R^n$ a curve. If I integrate $\omega$ along $\alpha$, I compute the scalar product of the
vectors (given by $\omega$) on $\alpha$ and the tangential vector that spans the tangential space of $\alpha$. (This is not quite precise because $\omega$ maps into the dual space of $\mathbb R^n$, but this is just supposed to visualize the concept of this integral). Does the integration of a $k$-form ($k$ > 1) over $M$, in a way, generalize the scalar product? What kind of integration is this?
4) What is the connection between the exterior derivative and the 'classical' higher-dimensional derivative?
Note that a submanifold of $\mathbb R^n$ is way more than just a subset of 
$\mathbb R^n$. 
EDIT1: I withdraw question 3): "Why does $\omega$ assign each $p∈M$ a different multilinear form?", because it follows basically by definition.
EDIT2: I think that it would have been smarter to let $\alpha$ go from $[a,b] \to \mathbb R^n$
 A: The answer to your first question comes from measure theory . The way integration is defined is  only possible for the forms which are of the same  length as the dimension.  Also , for me i think integration as a pairing of k forms and n-k forms .Thus then it makes sense only for the top forms.
For the third question , i think at first $\omega$ a k form as a section of the cotangent bundle of    k th exterior algebra of $\mathbb{R}^n$ which is the dual of the tangent bundle. If you are familiar with this definition , then it completely makes sense to have $\omega(p)$ as a multilinear form (in particular alternating) for each p .
A: Given a curve $\alpha : \mathbb{R} \to M$, I strongly prefer to think of "integration along $\alpha$" not in terms of using $\alpha$ to pushing the notion of integration on $\mathbb{R}$ over to a notion of integration on $M$, but instead the reverse: of pulling an integral on $M$ back to an integral over $\mathbb{R}$.
That is,
$$ \int_\alpha \omega = \int_{-\infty}^{\infty} \alpha^*(\omega) $$
No tangent spaces or tangent vectors are involved at all. $k$-dimensional surface integrals are the same; you pull back a differential form to $\mathbb{R}^k$ and use the familiar (oriented) notion of integration there.
And the familiar (oriented) notion of integration on $\mathbb{R}^k$ is the integral of a $k$-form; e.g. on $\mathbb{R}^3$ with the standard coordinates, integrands are things like $f(x,y,z) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z$.
The traditional definition
$$ \int_\alpha \omega = \int_{-\infty}^{\infty} \omega(\alpha'(x)) \, \mathrm{d} x $$
can be seen to be the same, as (where $\hat{e}_x$ means the standard tangent vector located at $x$)
$$ \omega(\alpha'(x)) \, \mathrm{d}x = \omega(\alpha_*(\hat{e}_x)) \, \mathrm{d}x= \alpha^*(\omega)(\hat{e}_x) \mathrm{d}x = \alpha^*(\omega) $$
but I find the introduction of tangent vectors makes things unnecessarily complicated, and obscures the central idea that the two curves $\mathbb{R}$ and $\alpha$ are (locally) basically the same thing, so we can do calculus on $\alpha$ in terms of calculus on $\mathbb{R}$.

So long as we think of a $k$-dimensional surface as comprising infinitesimal $k$-vectors at each point, an (oriented) $k$-dimensional volume operator ought to assign to each point a functional on $k$-vectors.
(and much of the technical complication is simply that standard analysis lacks an infinitesimal length scale)
Differential $k$-forms are simply the fields whose values are the infinitesimal representations of volume operators.
Since we are only retaining first-order information at the 'infinitesimal scale', everything (multi-)linearizes.

Differentials are first-order derivative information only. To talk about higher order differentiation, I think the right notion is via jets
