In my opinion, a lot of these relationships are suggested by abusive notation, abuses that hide what's really going on.
Don't get me wrong: some abuses of notation are harmless, or at the least, they help people get going on doing calculations. But they should still be understood to the fullest degree for those who wish to go beyond merely doing calculations.
I'll give an example: consider the relationship,
$$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$$
You probably know that differentials shouldn't really be divided, that this notation is really only suggestive, and while what it says is true by the inverse function theorem, it does so in a voodoo-like way that doesn't stand up to closer inspection, raising more questions than answers.
Of course, there's a totally reasonable way to phrase this notion: as I said, it's the inverse function theorem. Given a function $f$ on a vector $x$, we have the Jacobian $J_f$, and we know that
$$J_{f^{-1},f(x)} = J_{f,x}^{-1}$$
Which is a totally rigorous, though perhaps less obviously useful, statement.
(You might be thinking that nonstandard analysis could be useful here. Perhaps it would be, but my point is a bit larger: to understand and feel comfortable with the statement, you need to either take for granted that it stands in for something else, or accept that you need more math to understand it the way it's written.)
So, how does this relate to differentials and differential forms?
Well, mostly through the use of $d$ to denote the exterior derivative. Changing this symbol reveals how manifestly nonsensical some apparent relationships are.
For the purposes of this answer, I'll denote the exterior derivative by $\nabla$. This is reasonably familiar to students of vector calculus in 3d, and most of the results can be used directly from there.
Let's address your point (1), the total differential. It would be written as,
$$\nabla f = (\partial_i f) \nabla x^i$$
Again, recognizing the connection between the exterior derivative and the gradient from vector calculus, you should realize that the $\nabla x^i$ are nothing more than a set of basis vectors (more exactly, basis covectors), and all this does is decompose the gradient of $f$ into some coordinate directions. There is no explicit connection here between the gradient and differentials.
Let's talk about point (2), integrals around curves.
This is a common misconception from people who work with differential forms. I'll point out that the quantity $r'(t) = (x', y', z')(t)$ is manifestly a tangent vector. It literally points tangent to the curve that is the domain of integration, and fundamentally, it obeys quite different transformation laws than any form.
Moreover, if $F$ is a one-form, then it should be written
$$F = F_x \nabla x + F_y \nabla y + F_z \nabla z$$
If all the supposed $dx$'s are coming from the form, then what's coming from the $dl$? As argued above, what comes with $dl$ is not a set of basis forms but a vector, the tangent vector to the curve. Writing this vector $\ell'(t) = x' \partial_x r + y' \partial_y r + z' \partial_z r$ (where $r$ is a vector), we get for the dot product,
$$\int F \cdot dl = \int (F_x \circ l)(t) x'(t) \nabla x \cdot \partial_x r + \ldots \, dt$$
Of course, $\nabla x \cdot \partial_x r = 1$ by definition--otherwise, the basis forms would not be dual to the basis vectors. What would happen if we wrote the basis forms with the usual $dx$ notation?
$$\int F \cdot dl = \int (F_x \circ l)(t) x'(t) dx(\partial_x r) + \ldots \, dt$$
On its face, this looks like gobbledy-gook. Even if you had the presence of mind to distinguish between a basis form $dx$ and a differential denoting the variable of integration $dt$, it would be challenging to reconcile how these two notions should coexist in the same integral. I know I've met one person on this very site who suggested that no one should ever work with $dx$ and the like because you're just going to pull back anyway, so only $dt$ should be viewed as a differential form on this curve. That's...certainly one way of looking at things. To me, that comes at a high price of not being able to look at things geometrically. Let me explain:
What are you doing when you pull back a form in an integral like this? You're making it so the tangent vector in the target space has constant direction and magnitude (since you're pulling back to a 1d vector space, the image of the tangent vector is just the trivial unit vector). This is what's commonly done for form integrals, because then all your complexity is in the form, and in the Jacobian transforming that form, rather than in considering the components of the tangent vector. For this reason, the tangent vector is sometimes forgotten or neglected, since once you've pulled back, it's some trivial constant vector that will just be eaten by the form anyway. All that remains to be done is to set some convention for what direction it should be: positive or negative.
Anyway, you could call a basis form on that space by name, and perhaps some people would call it $dt$. If that abstract way of thinking works for you, do what you feel is best.
Finally, let's talk about point (3): this is more of a geometric interpretation question, and it's not unique to differential forms. Should a vector field be viewed as small, directed lines at every point? This is certainly behind the notion of field lines, which are commonly used for electric fields. I'm not sure I could say one (vectors) is more differential than the other (forms). Both involve orientations and magnitudes. In the end, I have to offer the same perspective as I would for vectors: does it make sense to think of a vector as a small piece of a line? If so, how would you decide that differentials are associated with forms instead of vectors? If not, how is this different from what you've done with forms?
Let me not digress for too long. There's a reason the notation for differential forms has stuck around as long as it has: it's enormously suggestive, and for dealing with unfamiliar concepts, suggestive notation is powerful. But like with the inverse function theorem, I submit that that notation is merely suggestive, full of shortcuts and sleight of hand. I do not think differential forms turn infinitesimals rigorous--far from it, I think that a far stronger relationship between forms and these differentials in integrals is suggested by the notation in ways that it shouldn't be.
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.
Best Answer
Let's work in $\Bbb R^n$ for simplicity (this is about intuition, after all).
Suppose $\omega = \sum_k\omega_k\mathrm d x^k$ is a nice $1$-form (covector field) on $\Bbb R^n$: then it can be converted into a vector field $\boldsymbol\omega := \omega^\sharp = \sum_k\omega_k \partial_{x^k}$ with the same components[1]. When you evaluate $\omega(\boldsymbol v)$ for some vector field $\boldsymbol v$ on $\Bbb R^3$, you obtain a scalar field; the meaning of this scalar field is clarified by "translating" $\omega$ into $\boldsymbol \omega$: with the Einstein summation convention, $$\omega (\boldsymbol v)=\omega_i\ \mathrm dx^i\left(v^j \frac{\partial}{\partial x^j}\right) = \omega_iv^i\equiv \boldsymbol \omega \cdot\boldsymbol v,$$ which is the scalar field obtained by projecting $\boldsymbol v$ onto $\boldsymbol \omega$ (or viceversa!) point by point. Thus, if you draw a small segment (a small $1$-parallelotope) at a specific $p\in\Bbb R^n$ in such a way that it is parallel to $\boldsymbol v$, in some sense, $\omega(\boldsymbol v)$ can be seen to represent a "(tangential) flux" along the segment. This is basically an infinitesimal interpretation of the line integral of $\boldsymbol \omega$ in vector calculus (ideally taken along a very short curve).
This intuition can be extended to $k$-degree forms, as long as you feed them $k$ vector fields instead of one; these fields then describe a $k$-parallelotope at all points.
The reasoning above is also why it is often said that $\omega$ are "made to be integrated": specifically, $k$-forms like to be integrated on manifolds of dimension $k$. If you imagine a regular enough $k$-manifold $M$ (embedded in $\Bbb R^n$, again for simplicity) and a $k$-form on it, then at each point $p\in M$ one could draw $k$ vectors $\boldsymbol v_{1p},\dots,\boldsymbol v_{kp}$ tangent to $M$, and if their modulus is taken to be smaller and smaller their $k$-parallelotope "approximates" $M$ better and better around $p$. As we know, the form $\omega$ describes a flux across every such parallelotope, and the integral over $M$ is a way to "compound" all these fluxes along $M$. In the case $k=1$, this represents the line integral from vector calculus (finite this time: think physical work of a force along a path); in the case $k=2$, this is a surface integral; and so on.
This digression on integrals comes in handy when thinking about the exterior derivative. If $\omega$ is a $k$-form, its exterior derivative $\mathrm d\omega$ is a $(k+1)$-form, and is naturally integrated along a regular $(k+1)$-manifold $M$. Stokes' theorem is the miraculous fact that, if this manifold has a boundary and is nice enough[2], then $$\int_M \mathrm d\omega = \int_{\partial M}\omega. $$ (This is a generalization of the gradient, curl, and divergence theorems from vector calculus.) Now suppose you take $M$ and "contract it" around a point $p\in M$: then it looks more and more like a $(k+1)$-parallelotope at $p$, and the action of $\omega$ can again be interpreted as a flux of $\mathrm d\omega$ along this parallelotope. But by Stokes' theorem, this corresponds to the total flux of $\omega$ along the boundary faces of the parallelotope (which are $k$-parallelotopes)!
Footnotes. [1] This is not something we can do in all manifolds: the musical isomorphisms $(-)^\sharp$ and $(-)^\flat$ are only defined in manifolds where there is a (pseudo)metric tensor $g$, i.e. (pseudo-)Riemannian manifolds. Here, we are thinking of $\Bbb R^n$ as a Riemannian manifold with the standard Euclidean metric $g(\partial_{x^i},\partial_{x^k})=\delta_{ik}$.
[2] Since a $k$-form is alternating in its arguments, it matters in what order you feed it vector fields. Thus, when approximating a manifold $M$ by $k$-parallelotopes, you need to make sure that the parallelotopes "vary smoothly" along the surface (i.e. the vector fields $\boldsymbol v_i$ describing their edges must be smooth) and that, if you follow a closed loop from a point $p\in M$ to itself, you don't unknowingly switch some of the edges around. Not all manifolds can be "traced out" by vector fields with these properties; a famous counterexample is the Möbius strip (try it!). When the manifold does admit these edge fields, it is said to be orientable, and a choice of edge fields is said to define an orientation. Once you calculate an integral in a fixed orientation, switching two of the edge fields and calculating again leads to minus the integral you had before!
Stokes' theorem holds only for orientable manifolds with boundary, and relies on the fact that, once an orientation is chosen on $M$, it induces a natural orientation on $\partial M$—which is the one with respect to which the integral on the LHS needs to be calculated; otherwise, you get a pesky sign error.