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 $V$!
I don't think that there is a uniform use of terminology, although it seems more common to me that "vector valued differential forms" refers to the case of a fixed vector space, whereas in the other case, I would use the terminology "vector bundle valued differential forms". An important difference between the two cases is that in the former case, there is a natural version of the exterior derivative, while in the latter case, one has to choose a linear connection on the vector bundle in order to have an analog of the exterior derivative.
The case of $\mathfrak g$-valued forms on a principal bundle you mention in your comment is the case of a fixed vector space (and if you define connection forms in that way, you need the exterior derivative on vector valued forms in the definition of curvature). Principal bundles also provide a nice relation between the two cases: Suppose $P\to M$ is a principal $G$-bundle, $V$ is a representation of $G$ (so in particular a finite dimensional vector space). Then there is a the associated vector bundle $E=P\times_G V\to M$ and one can identify the space $\Omega^*(M,E)$ of vector bundle valued forms with a subspace of the space $\Omega^*(P,V)$ of vector valued differential forms. The elements in that subspace are exactly those forms, which are horizontal and $P$-equivariant.
Best Answer
Note: I figured out how to do this from McDuff's J-Holomorphic Curves book. Adding some details here in order to close the question.
In order to formalize the differential operator in a global setup, consider the space $\mathcal{B}=C^\infty(\Sigma, M)$. For $f:\Sigma\to M$, denote the space $\mathcal{E}_f=\Gamma \big(f^*E\otimes \Lambda^pT^*\Sigma\big)$. Then we can think of a bundle $\mathcal{E}\to\mathcal{B}$. This is an infinite dimensional vector bundle, where the spaces have Frechet topology. The operator $$D:f\mapsto f^*\omega$$ can then be realized as a section of this bundle. This seems like the correct way to study the operator.
As an aside, one can get the linearization operator associated to $D$ at $f$, by considering some arbitrary connection $\nabla$ on $E$. One then gets the operator, $$\begin{align*}L_f : \Gamma f^*TM &\to \Gamma\hom(\Lambda^p TV, f^*E) \\ \xi &\mapsto \mathfrak{L}_\xi\omega\end{align*}$$ Here $\mathfrak{L}$ is the Lie derivative of $\omega$ along $\xi$ and is given by the Cartan formula, $\mathfrak{L}_\xi\omega = \iota_\xi d_\nabla\omega + d_\nabla\iota_\xi\omega$. The details for Lie derivative by a vector field along a map, may found in Michor's book.