Let $E$ be a vector bundle over $M$. Wikipedia says an $E$-valued differential form over $M$ can be defined as a bundle morphism $TM\otimes...\otimes TM \rightarrow E$ that is skew-symmetric. I am able to make the identification of this with $\Gamma (E^*)\otimes\Omega^p(M)$ as $\alpha_p(\omega,X_{1},X_{2},..)=\omega_p(\alpha_p(X_{1p},X_{2p},..))$.
Isn't it supposed to be $\Gamma(E\otimes\Omega^p(M))$? Feel like I am missing/misunderstanding something very straightforward. What am I doing wrong?
Let $V,W$ be vector spaces. By relativizing the natural map $V^\vee\otimes W\to \mathsf{Vect}(V,W)$ (an isomorphism in the finite dimensional case) to vector bundles $V,W$ over a manifold, I believe we get an isomorphism between the sections of the vector bundle $V^\vee\otimes W$ and vector bundle morphisms $V\to W$.
A skew-symmetric bundle map as you describe in your answer is equivalently a bundle map $\Lambda^n\mathrm TX\to E$. By the previous paragraph this amounts to a section of $(\Lambda ^n\mathrm TX)^\vee\otimes E$. Now, the perfect determinant pairing induces a canonical isomorphism $(\Lambda^n\mathrm TX)^\vee\cong \Lambda^n(\mathrm T^\vee X)$. The latter is the bundle of differential $n$-forms. Thus the pairing gives a canonical isomorphism $(\Lambda ^n\mathrm TX)^\vee\otimes E\cong \Omega ^n_X\otimes E$.
One way to illustrate why the $E$ remains undualized is as follows. A bundle morphism $$ TM^{\otimes n}\to E $$ is the same thing as a bundle morphism $$ TM^{\otimes n-1}\to E\otimes T^*M, $$ which is the same thing as a bundle morphism
$$ TM^{\otimes n-2}\to E\otimes T^*M^{\otimes 2}, $$ and so on until we are left with
$$ TM\to E\otimes T^*M^{\otimes n-1}, $$ and finally a section of $E\otimes T^*M^{\otimes n}$. Enforcing antisymmetry all along gives you the identification you seek.
In summary, since $E$ never "changes sides" of the mapping, it never gets dualized.
