Ways of thinking about vector-valued differential forms I am trying to get a better intuition of vector-valued differential forms.
Let $V$ be a vector space and $M$ a smooth manifold. Consider the space $\Omega^k(M;V)=\Gamma((M\times V)\otimes \Lambda^k(T^*M))$. That is, the space of $V$ valued $k$-forms on $M$. 
Intuitively I think of this definition as: For each $x\in M$, $\omega(x)$ is an alternating multi-linear map that maps $k$ tangent vectors at $x$ to a vector $v\in V$. In terms of the precise definition:
For $x\in M$, if $\omega\in \Omega^k(M;V)$, we have
$\omega(x)\in (\pi^{-1}(x))\otimes \{x\}\times\Lambda^k(T^*_xM)$
Firstly - is the above line correct? Since $\Lambda^k(T^*M)$ is a disjoint union, I would think that a section of $\Lambda^k(T^*M)$ would map an element $x$ of $M$ to $\{x\}\times T^*_xM$. So, because of the $\{x\}$ in the above equation, I feel like it is much more natural to think of $\omega(x)$ as the map described above but that it also outputs the point $x$ as well.
That is, $\omega(x)(X_1,\dots,X_k)=(x,v)$. Is this wrong? 
What makes me a little uneasy is that since the bundle we are considering is trivial, does it not follow that $\pi^{-1}(x)$ is precisely $\{x\}\times V$? So then should it be
$\omega(x)\in (\{x\}\times V)\otimes (\{x\}\times \Lambda^k(T^*_x M))$...
Which just looks weird...
My understanding is not quite there yet. If anyone could help with this, it would be much appreciated.
 A: For ease of notation, let $(\mathbb{V}, \pi)$ be the trivial vector bundle $M \times V$; i.e., its total space is $M \times V$, and $\pi: M \times V \to M$ is projection onto the first factor. You are correct that $\pi^{-1}(x) = \{x\} \times V$, but this is canonically identifiable with $V$, so it's most natural to think of $\pi^{-1}(x)$ as simply $V$.
Let $\tilde{\pi}: \Lambda^k (T^\ast M) \to M$ be the projection. The fiber $\tilde{\pi}^{-1}(x)$ is usually denoted simply by $\Lambda^k (T_x^\ast M)$ (I think your $\{x \}$ is extraneous). By definition this is the vector space of alternating maps mapping $k$ tangent vectors in $T_x M$ to $\mathbb{R}$.
We are interested in the vector bundle $\mathbb{V} \otimes \Lambda^k (T^\ast M)$. Let's call it $(E, \pi_E)$, with $\pi_E: E \to M$. The tensor product means that the fiber over $x \in M$ is the vector space tensor product of the fibers of $\mathbb{V}$ and $\Lambda^k (T^\ast M)$:
\begin{align*}
 E_x :=& \pi_E^{-1}(x) \\
      :=& \pi^{-1}(x) \otimes \tilde{\pi}^{-1}(x) \\
      =& V \otimes \Lambda^k (T_x^\ast M)
\end{align*}
So for a section $\omega$ of $E$, $\omega(x) \in E_x = V \otimes \Lambda^k (T_x^\ast M)$. As you say, one way to think about this is as an alternating map mapping $k$ tangent vectors to a vector in $V$: $\omega(x)(X_1, \dots, X_k) \in V$. (Again, I think your $x$'s are extraneous.)
Here's a way to think of vector-valued differential forms that may feel more concrete. Fix a basis $\{v_1, \dots, v_n\}$ for $V$. A vector $v$ in $V$ can then be specified by an $n$-tuple of real numbers $(a_1, \dots, a_n)$ (where $v = \sum_{j=1}^n a_j v_j$). Similarly, an element $\omega(x)$ of $E_x = V \otimes \Lambda^k (T_x^\ast M)$ can be specified by an $n$-tuple of $k$-forms $(\alpha_1, \dots, \alpha_n)$, where each $\alpha_j \in \Lambda^k (T_x^\ast M)$ and $\omega(x) = \sum_{j=1}^n v_j \otimes \alpha_j$. Since $\mathbb{V}$ is globally trivial, we can use the same basis for $V$ at every point, and think of a global section $\omega$ of $\mathbb{V}$ as an $n$-tuple of differential forms $(\omega_1, \dots, \omega_n)$, where each $\omega_j \in \Gamma(\Lambda^k (T^\ast M))$ and at each point $x$, $\omega(x)=\sum_{j=1}^n v_j \otimes \omega_j(x)$. So a vector-valued differential form is really just specified by $n$ (ordinary) differential forms, once we fix a basis for $V$!
