A $\frak g$-valued differential form is , as far as I know, just a section $\alpha$ of the tensor product of the exterior power of the cotangent bundle $\Lambda^{\bullet}T^*M$ of some manifold $M$ with the trivial vector bundle $M\times\frak{g}$. As such, locally over some chart domain $U$, $\alpha$ can be cast in the follwing form
$$\alpha\equiv\alpha_1\otimes x_1+\cdots+\alpha_n\otimes x_n$$
where $\alpha_1,\dots,\alpha_n$ are local differential forms on $M$ defined over the chart domain $U$, and $x_1,\dots,x_n$ is a basis of $\frak g$. The differential is then calculated by ignoring the Lie algebra terms:
$$d\alpha\equiv (d\alpha_1)\otimes x_1+\cdots+(d\alpha_n)\otimes x_n$$
Similarly, the product is defined by treating the differential forms and the Lie algebra elements as separate entities:
$$[\alpha\wedge\beta]=\sum_{1\leq i,j\leq n}\alpha_i\wedge\beta_j\otimes[x_i,x_j]$$
For instance, for a pure form $\alpha$ of degree $p$, what you know about the exterior differential immediately implies that
$$d[\alpha\wedge\beta]=[(d\alpha)\wedge\beta]+(-1)^p[\alpha\wedge(d\beta)]$$
Also, if $\alpha$ has degree $p$, and $\beta$ has degree $q$, then
$$[\beta\wedge\alpha]=(-1)^{pq+1}[\alpha\wedge\beta]$$
I think the algebraic questions that arise are easy enough that I'm sure you can find all the relations you want on your own. However, you can always take a look at Peter W. Michor's Topics in Differential Geometry, in particular his chapter IV, ยง19, or Morita's Geometry of Characteristic Classes.
The overwhelming preference to work with $k$-covector fields ("differential forms") stems from a few basic facts:
First, you might know of $\nabla$ from vector calculus. It is related to the exterior derivative $d$ in the sense that you can do $\nabla \wedge$ on a covector field and it is equivalent to $d$. $\nabla$ itself transforms as a covector does, and so it takes 1-covectors to 2-covectors, $k$-covectors to $k+1$-covectors (and these are all fields, of course). So there is a very convenient element of closure under the operation.
Second, integration on a manifold naturally involves the tangent $k$-vector of the manifold. This is something traditional differential forms notation tends to gloss over. When you see, for example, something like this:
$$\int f \, \mathrm dx^1 \wedge \mathrm dx^2$$
It really means this:
$$\int f \, (\mathrm dx^1 \wedge \mathrm dx^2)(e_1 \wedge e_2) \, dx^1 \, dx^2$$
For this reason, the basis covectors $\mathrm dx^i$ should not be confused with the differentials $dx^i$. Further, that we use $e_1 \wedge e_2$ here, and not $e_2 \wedge e_1$, reflects an implicit choice of orientation, which is usually picked by convention from the ordering of the basis, but this need not always be the case. The tangent $k$-vector, and especially its orientation, must necessarily be considered in these integrals.
So why does this make $k$-covector-fields preferred? Because the action of these fields on the manifolds' tangent $k$-vectors is inherently nonmetrical. So, differential forms allows you to do a lot of calculus without imposing a metric.
This point, however, is somewhat obfuscated when you introduce the Hodge star operator and interior differentials, for these are metrical. Then, you get a big problem with differential forms: by working exclusively with $k$-covector fields, and expunging all reference to $k$-vector fields, the treatment when we do have a metric is extremely ham-fisted. Yes, you can do everything with wedges, exterior derivatives, and Hodge stars. But it makes much more sense to use corresponding grade-lowering operations and derivatives instead. Geometric calculus does this, but let met get to that in a moment.
Regarding the pushforward vs. pullback, I must confess a lack of understanding. I do not see why we would want to pull covectors back from a target manifold while insisting we must push vectors forward. I'm very familiar with the mathematics: that under a smooth map, the adjoint Jacobian transforms covectors from the target cotangent space to the original, and the inverse Jacobian does the same for vectors. Perhaps it has to do with defining the pushforward as the inverse of this inverse.
Now, do all these remarks put together mean that $k$-vector fields are inherently disadvantaged, or less rich, than $k$-covector fields? I would say no. I mentioned geometric calculus earlier: it is the originator of the $\nabla \wedge$ notation that I used earlier, and it handles $k$-vector fields just fine. Geometric calculus is the calculus that goes with clifford algebra, and you may find it illuminating. Many of the theorems and results of differential forms translate to geometric calculus and to $k$-vector fields. Stokes' theorem? Used extensively. de Rham cohomology? Most of the same results apply.
My point above about differential forms integrals using tangent $k$-vectors implicitly? That comes from geometric calculus, too, where the tangent $k$-vector is not digested "trivially" and you have to look at all the metrical ways in which it might interact with the vector field you're integrating.
A grade-lowering derivative is natural to use with $k$-vector fields. In geometric calculus, this is notated as $\nabla \cdot$. You can see that successive chains of $\nabla \cdot$ continually lower the grade of a field, just as successive exterior derivatives raise it.
My ultimate point is that, when you do have a metric, it's quite nonsensical to treat everything as a differential form instead of using $k$-vector fields when appropriate. I feel the tendency to do this in physics divorces students from a lot of the vector calculus they had learned, unnecessarily so. I can't speak to mathematics courses, but I imagine some of that criticism applies, too.
Now, there are some properties of covector fields and exterior derivatives that are nicer than working with vector fields. For instance, under a map $f(x) = x'$ with adjoint Jacobian $\overline f$, it's true that $\overline f(\nabla ' \wedge A') = \nabla \wedge A$ for some covector field $A$. That's a very convenient result, and there's no correspondingly nice identity for vector fields.
Best Answer
Looking through the slides you linked, it looks like they can all be done without working in coordinates. Do you perhaps have specific identities in mind for which you mean ' possible to argue without coordinates.' You mentioned Cartan's magic formula can be proved without working in local coordinates, and so can $\mathcal L_{[X,Y]}\alpha=\mathcal L_X(\mathcal L_Y\alpha)-\mathcal L_Y(\mathcal L_X\alpha)$ as it is just the Jacobi identity for a Lie-bracket.
For your second question, all of those identities can be found in most introduction to differential geometry books. Check out John Lee's "Introduction to Smooth Manifolds" and Loring Tu's "An Introduction to Manifolds." The identities will either be derived or left as an exercise. But those left as an exercise can probably be found in another introduction to differential geometry book.