Flatness as an integrability condition without invoking bundles

Let $$\pi:E\rightarrow M$$ be a vector bundle, and let $$U\subseteq M$$ be a trivialization domain for $$E$$. Assume a linear connection is given on $$E$$ with local connection form(s) $$\omega=(\omega^a_{\ b})$$.

Let $$\psi\in\Gamma_E(U)$$ be a local section, with components in the trivialization being $$\psi^a$$.

The differential equation of $$\psi$$ being absolute parallel is $$0=(d_\omega\psi)^a=d\psi^a+\omega^a_{\ b}\psi^b.$$

I can use Frobenius' theorem to show that the integrability condition for this equation is $$0=\Omega=d\omega+\omega\wedge\omega$$ by going to the total space $$E$$. If local coordinates on $$E$$ are $$(x^\mu,u^a)$$, then the horizontal distribution is locally given by $$\mathcal H=\cap_a\ker(\delta u^a)$$ with $$\delta u^a=du^a+\omega^a_{\ b}u^b,$$ where $$\omega$$ here is understood as a tuple of 1-forms on $$E$$ via the canonical horizontal isomorphism. Then clearly $$(d_\omega\psi)^a=\psi^\ast\delta u^a$$, and if the equation is satisfied, then $$\psi$$ as a map from $$M$$ to $$E$$ must essentially define a horizontal submanifold of $$E$$, hence we know that the Frobenius integrability of $$\mathcal H$$ is equivalent to the solvability of the equation.

We know from Frobenius that the horizontal distribution is integrable if and only if the differentials of $$\delta u^a$$ also annihilate the horizontal. Calculating the differential gives $$d\delta u^a=du^b\wedge\omega^a_{\ b}+d\omega^a_{\ b}u^b.$$ Since we are restricting horizontally, it is true that $$du^b=-\omega^b_{\ c}u^c$$, so reinserting this gives $$0=d\delta u^a|_\mathcal H=\omega^a_{\ b}\wedge\omega^b_{\ c}u^c+d\omega^a_{\ b}u^b.$$ Renaming indices, factoring out the fiber coordinates and demanding that this result should be true for all of them (and realizing that the expression trivially annihilates verticals) gives the integrability condition $$d\omega+\omega\wedge\omega=0.$$

However, I'd like to get this integrability condition without working on the total space at all. In this case, I don't know how to use Frobenius, because as someone who is familiar with differential geometry, but not very familiar with PDEs, I can only apply Frobenius to a tangent distribution, and if I only have $$0=d\psi^a+\omega^a_{\ b}\psi^b,$$ a local equation on the base space, I don't see a distribution I can apply Frobenius to.

Question: This question is somewhat motivated by pedagogy. What is the most pedagogical, least "abstract" and absolutely not bundle-requiring method to show that the integrability condition is $$d\omega+\omega\wedge\omega=0$$?

Consider the manifold $$X=M\times G$$ (where $$G=GL(n)$$ for simplicity). Define a distribution $$\mathcal H$$ on $$X$$ as the image of your connection 1-form, i.e.: $$\mathcal H_{p,g}:=\{ (X_p,\omega_p(X) g)\mid X\in T_pM\}\subset T_pM\times TgG=T_{(p,g)}X.$$ The integrability condition of $$\mathcal H$$ is just flatness $$dw+\omega\wedge\omega=0$$, so you can apply Frobenius here. As an integral manifold of $$\mathcal H$$ locally projects to the first factor, you can regard an integral manifold locally as a map $$g\colon U\subset M\to G$$ satisfying $$\omega=dg g^{-1}.$$