I have several answers which illustrate the ideas in many special cases (but once you ‘get’ the idea, you can do almost anything you want). Besides, you should indeed follow a textbook in detail rather than doing small pieces here and there because otherwise you’ll always be confused when a new situation pops up.
I’m sure there are more, but I’m too lazy to find them all. Anyway, here’s a consolidation of the ‘basics’.
1. Vector Space level.
The most important thing to understand is ALWAYS at the vector space level. Let $V,E$ be a finite-dimensional (real) vector spaces. Then, basic (multi)linear algebra tells us that the following are equivalent in terms of ‘information’:
- an alternating multilinear map $\underbrace{V\times\cdots\times V}_{\text{$k$ times}}\to E$ (a common notation for the space of such maps is $\mathcal{A}^k(V,E)$).
- a linear map $\bigwedge^k(V)\to E$, i.e an element of the vector space $\text{Hom}\left(\bigwedge^k(V),E\right)$
- an element of $\bigg[\bigwedge^k(V)\bigg]^*\otimes E$, which is simply written as $\bigwedge^k(V)^*\otimes E$
- an element of $\bigg[ \bigwedge^k(V^*)\bigg]\otimes E$, which is simply written as $\bigwedge^k(V^*)\otimes E$
The meaning of the $\otimes$ is that of a general tensor product of vector spaces: for any two vector spaces $X,Y$ one can construct a new vector space $X\otimes Y$ which satisfies a certain universal property (roughly, it amounts to taking ‘formal linear combinations’ of the symbols $x\otimes y$ for $x\in X,y\in Y$). This is covered in any good algebra textbook, so I won’t repeat it all here.
The equivalence of the first two is by the universal property of exterior powers (i.e almost by definition). The equivalence of the second and third is the usual Hom/tensor-product isomorphism that for finite-dimensional vector spaces $W,E$, $\text{Hom}(W,E)\cong W^*\otimes E$ (the way I think of this is that $W^*=\text{Hom}(W,\Bbb{R})$ consists of real-valued linear maps, so if we want $E$-valued linear maps, we need to ‘enrich the target space’, and this is done by tensor products). The equivalence of the third and fourth descriptions is by basic algebra involving duals and how it interacts with exterior powers. I’ll let you google these facts. Also, all these isomorphisms are canonical, so algebraically speaking, no one description is inherently better than any other.
Pedagogically, the first requires the least amount of preliminary vocabulary, so in that sense, it is the ‘simplest’. However, ‘easy to present’ doesn’t always mean ‘easy to work with’ or ‘quick to work with’. Once you gain familiarity with tensor products and exterior powers, you’ll see that you can do various things rapidly (tensor products of two maps $f_1:V_1\to E_1$ and $f_2:V_2\to E_2$ to get $f_1\otimes f_2:V_1\otimes V_2\to E_1\otimes E_2$, and similar stuff for exterior powers), so the other descriptions are also good to know. So, the fact that at the vector space level we have (at least) 4 ways of describing the same thing, it is no surprise that at the vector-bundle level, you get more ways/notations all of which are trying to say the same thing.
2. Vector Bundle stuff
Let $E,F$ be smooth vector bundles over $M$. The following objects convey the same information:
- a smooth vector bundle morphism $\Phi:E\to F$ (i.e a smooth map $\Phi:E\to F$ such that for each $x\in M$, $\Phi$ restricts to a linear map of the fiber $E_x$ into the fiber $F_x$).
- a smooth section $\tilde{\Phi}$ of the vector bundle $\text{Hom}(E,F)$ over $M$.
- a smooth section $\tilde{\tilde{\Phi}}$ of the vector bundle $E^*\otimes F$ over $M$
The equivalence of the first and second is simply by setting $\tilde{\Phi}(x)=\Phi|_{E_x\to F_x}$. Proving that smoothness of $\Phi$ is equivalent to smoothness of $\tilde{\Phi}$ is done writing things out in local vector-bundle charts. The equivalence of the second and third is by the linear-algebra analogue (done fiberwise). Symbolically, you could write this as
\begin{align}
\text{Mor}(E,F)\cong \Gamma\left(\text{Hom}(E,F)\right)\cong\Gamma(E^*\otimes F).
\end{align}
3. Putting things together
Let $E$ be a vector bundle over $M$ and $k\geq 0$ an integer. In view of the vector-space discussion and and above vector bundle discussion, we can now define a exterior differential $k$-form on $M$ with values in $E$ to be any of the following objects:
- a smooth fiberwise-alternating multilinear morphism $(TM)^{\oplus k}\to E$
- a smooth vector bundle morphism $\bigwedge^k(TM)\to E$
- a smooth section of the vector bundle $\text{Hom}\left(\bigwedge^k(TM), E\right)$
- a smooth section of the vector bundle, $\left[\bigwedge^k(TM)\right]^*\otimes E$
- a smooth section of the vector bundle $\left[\bigwedge^k(T^*M)\right]\otimes E$
Regardless of which ‘concrete’ implementation you decide to take, the space of all such $E$-valued $k$-forms on $M$ is often denoted simply $\Omega^k(M;E)$, and when $E=M\times V$ is a trivial vector bundle (with $V$ the common vector space), then we don’t write $\Omega^k(M; M\times V)$ but rather simply $\Omega^k(M;V)$. Finally, when $V=\Bbb{R}$ (i.e the usual (scalar) differential $k$-forms on $M$) we write $\Omega^k(M)$ rather than $\Omega^k(M;\Bbb{R})$.
