Why does the tensor product appear in the codomain of a Lie Algebra valued one-form?

I'm watching a lecture series on differential geometry and a Lie-algebra valued one-form $$A$$ on a principle $$G-$$bundle $$(P, \pi, M)$$ is written to belong to the space $$A \in \Omega^1(M) \otimes T_e G$$ Where $$\Omega^1(M)$$ is the space of one-forms over $$M$$ and $$T_e G$$ is the lie algebra of the lie group $$G$$.

Can someone explain why the tensor product shows up here?

Remember from linear algebra that when working with finite-dimensional vector spaces (bundles), the space (bundle) of linear maps $$L(V, W)$$ is isomorphic to the tensor product $$V^* \otimes W.$$ Since the usual ($$\mathbb R$$-valued) differential one-forms are sections of the bundle $$TM^* = L(TM, \mathbb R),$$ we thus see that if $$E$$ is some other vector space or bundle, the $$E$$-valued one-forms on $$M$$ should be sections of $$L(TM,E) = TM^* \otimes E.$$
In this particular case, $$E$$ is the vector space $$T_e G$$ (the Lie algebra of $$G$$), so the $$E$$-valued one-forms are sections of $$TM^*\otimes T_e G.$$ (If you want to get pedantic about this being a tensor product of vector bundles over $$M$$, $$T_e G$$ here is really shorthand for the trivial bundle $$M \otimes T_e G \to M.$$)
The space of sections of this bundle is $$\Gamma(TM^* \otimes T_e G) = \Gamma(TM^*) \otimes T_e G = \Omega^1(M) \otimes T_eG.$$
A one-form eats tangent vectors and returns scalars, so a $$G$$-valued one-form ought to eat tangent vectors and return elements of $$G$$.
An element of $$\Omega \otimes G$$ does exactly this, viewing $$\omega \otimes g$$ as the thing that eats $$v$$ and returns $$\omega(v) \cdot g$$.