I've read that we form the exterior algebra $\Omega^*(M)$ of differential forms on a smooth $n$-dimensional manifold $M$ by $$\Omega^*(M) = \bigoplus_{k=0}^n \Omega^k(M),$$ so it is a direct sum of the spaces of differential $k$-forms for all $k$. What if we approached this from a slightly different angle, and created a vector bundle over $M$ where the fiber over $p \in M$ is $\Lambda(T^*_pM)$ (so we create the graded algebras over each point first and then sew them together, instead of vice versa.) Will this create the same object?
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Your proposed construction is consistent to taking the direct sum of all the differential $k$-forms. Note to construct $\Lambda(T_p^*M)$ we need to construct each fiber as a Whitney sum anyway:
$$ \Lambda(T_p^*M) \;\; =\;\; \bigoplus_{k=0}^n \Lambda^k(T_p^*M). $$
I suppose the one key difference is to just remember the difference between the graded alternating tensor bundle $\Lambda(T^*M)$ and differential forms $\Omega^*(M)$. Elements $\sigma \in \Omega^*(M)$ are smooth sections $\sigma: M \to \Lambda(T^*M)$.
