I have struggled with this question myself, and I couldn't find a perfectly satisfactory answer. In the end, I decided that the definition of a differential form is a rather strange compromise between geometric intuition and algebraic simplicity, and that it cannot be motivated by either of these by itself. Here, by geometric intuition I mean the idea that "differential forms are things that can be integrated" (as in Bachmann's notes), and by algebraic simplicity I mean the idea that they are linear.
The two parts of the definition that make perfect geometric sense are the d operator and the wedge product. The operator d is simply that operator for which Stokes' theorem holds, namely if you integrate d of a n-form over an n+1-dimensional manifold, you get the same thing as if you integrated the form over the n-dimensional boundary.
The wedge product is a bit harder to see geometrically, but it is in fact the proper analogy to the product measure. Here's how it works for one-forms. Suppose you have two one-forms a and b (on a vector space, for simplicity). Think of them as a way of measuring lengths, and suppose you want to measure area. Here's how you do it: pick a vector $\vec v$ such that $a(\vec v) \neq 0$ but $b(\vec v) = 0$ and a vector $\vec w$ s.t. $a(\vec w) = 0$ but $b(\vec w) \neq 0$. Declare the area of the parallelogram determined by $\vec v$ and $\vec w$ to be $a(\vec v) \cdot b(\vec w)$. By linearity, this will determine area of any parallelogram. So, we get a two-form, which is in fact precisely $a \wedge b$.
Now, the part that makes no sense to me geometrically is why the hell differential forms have to be linear. This implies all kinds of things that seem counterintuitive counter-intuitive to me; for example there is always a direction in which a one-form is zero, and so for any one-form you can draw a curve whose "length" with respect to the form is zero. More generally, when I was learning about forms, I was used to measures as those things which we integrate, and I still see no geometric reason as to why measures (and, in particular, areas) are not forms.
However, this does make perfect sense algebraically: we like linear forms, they are simple. For example (according to Bachmann), their linearity is the thing that allows the differential operator d to be defined in such a way that Stokes' theorem holds. Ultimately, however, I think the justification for this are all the short and sweet formulas (e.g. Cartan's formula) that make all kinds of calculations easier, and all depend on this linearity. Also, the crucial magical fact that d-s, wedges, and inner products of differential forms all remain differential forms needs this linearity.
Of course, if we want them to be linear, they will be also signed, and so measures will not be differential forms. To me, this seems as a small sacrifice of geometry for the same timesake of algebra. Still, I don't believe it's possible to motivate differential forms by algebra alone. In particular, the only way I could explain to myself why take the "Alt" of a product of forms in the definition of the wedge product is the geometric explanation above.
So, I think the motivation and power behind differential forms is that, without wholly belonging to either the algebraic or geometric worlds, they serve as a nice bridge in between. One thing that made me happier about all this is that, once you accept their definition as a given and get used to it, most of the proofs (again, I'm thinking of Cartan's formula) can be understood with the geometric intuition.
Needless to say, if anybody can improve on any of the above, I'll be very grateful to them.
P.S. For the sake of completeness: I think that "inner products" make perfect algebraic sense, but are easy to see geometrically as well.