I want to start with a simple analogy to the (ordinary) derivative. So suppose that
$\omega$ is a $k$-form, and $X_1, \ldots, X_k$ are vector fields. And for the moment, I want you to imagine that the $X_i$ fields are all "constant near some point $p$. Now that doesn't really make sense (unless you're in $\Bbb R^n$) but bear with me. If $p$ is the north pole, and $X_1(p)$ is a vector field that points towards, say, London, then it makes sense to define $X_1$ near $p$ to also point towards London, and those vectors will (in 3-space) all be pretty close to $X_1(p)$.
Then we can define a function
$$
f(q) = \omega(q)[X_1(q), \ldots, X_k(q)]
$$
defined for $q$ near $p$.
How does $f(q)$ vary as $q$ moves away from $p$? Well, it depends on the direction that $q$ moves. So we can ask: What is
$$
f(p + tv ) - f(p)?
$$
Or better still, what is
$$
\frac{f(p + tv ) - f(p)}{t}?
$$
especially as $t$ gets close to zero?
That "derivative" is almost the definition of
$$
d\omega(p)[X_1(p), \ldots, X_k(p), v].
$$
There are a couple of problems with that "definition" as it stands:
- What if there are multiple ways to extend $X_i(p)$, i.e., what if "constant" doesn't really make sense? Will the answer be the same regardless of the values of $X_i$ near $p$ (as opposed to at $p$)?
- How do we know that $d\omega$ has all those nice properties like being antisymmetric, etc.?
- How does this fit in with div, grad, curl, and all that?
Problems 1 and 2 are why we have fancy definitions of $d$ that make theorems easy to prove, but hide the insight. Let me just briefly attack item 3.
For a 0-form, $g$, the informal definition I gave above is exactly the definition of the gradient. You have to do some stuff with mixed partials (I think) to verify that the gradient, as a function of the vector $v$ is actually linear in $v$, and therefore can be write $dg(p)[v] = w \cdot v$ for some vector $w$, which we call the "gradient of $g$ at $p$."
So that case is pretty nice.
What about the curl? That one's messier, and it involves the identification of every alternating 2-form with a 1-form (because $2 + 1 = 3$), so I'm going to skip it.
What about div? For the most basic kind of 2-form, something like
$$
\omega(p) = h(x, y, z) dx \wedge dy
$$
and the point $p = (0,0,0)$ and the vector $v = (0,0,1)$, and the two "vector fields" $X_1(x,y,z) = (1,0,0)$ and $X_2(x, y, z) = (0, 1, 0)$, we end up looking at
\begin{align}
f(p + tv) &= h(0, 0, t) dx \wedge dy[ (1,0,0), (0, 1, 0)]\\
&= h(0,0, t)
\end{align}
and the difference quotient ends up being just
$$
\frac{\partial h}{\partial z}(0,0,0)
$$
That number tells you how $\omega's$ "response" to area in the $xy$-plane changes as you move in the $z$ direction.
What's that have to do with the divergence of a vector field? Well, that vector field is really a 2-form-field, and duality has been applied again. But in coordinates, it looks like $(0,0,h)$, and its divergence is exactly the $z$-derivative of $h$. So the two notions match up again in this case.
I apologize for not drawing out every detail; I think that the main insight comes from recognizing the idea that the exterior derivative is really just a directional derivative with respect to its last argument...and then doing the algebra to see that it's also a directional derivative with respect to the OTHER arguments as well, which is pretty cool and leads to cool things like Stokes' theorem.