I have some confusion regarding how the covariant derivative is defined for one forms on a manifold in the context of frames/vielbeins. I am a physics student and my reference is Sec 4.3 of the Lectures on Advanced Mathematical Methods for Physicists by Sunil Mukhi. I have given the context behind my confusion below.
Suppose we have some notion of orthonormal frames $E_a = E_a^{~\mu}~\partial_{\mu}$ on an $n$ dimensional manifold $\mathcal{M}$. Here, the Greek indices are coordinate indices and the Latin indices are indices with respect to $O(n)$ which "rotates" the frame $$E_a \to E'_a = M_a^{~b}~E_b~,$$ where $M_a^{~b} \in O(n)$.
Given such frames, one can also define "dual frames" $e^a = e^a_{~\mu}~dx^{\mu}$ such that $$\left<e^a,E_b\right> = \delta^a_{~b}$$
Now, suppose I have a one-form $A$ which has a coordinate representation $A = A_{\mu}~dx^{\mu}$. I can project this onto a frame of my choice to obtain $A_a = E_a^{~\mu}A_{\mu}$ which is a coordinate scalar but is still a vector under $O(n)$ i.e. $A_a \to A'_a = M_a^{~b}A_b$. If I now consider the exterior derivative, in the absence of the frames I have
$$dA = \partial_{\mu}A_{\nu}~dx^{\mu}\wedge dx^{\nu}~,$$
which is a two-form under coordinate transformations. However, $dA$ no longer transforms covariantly under $O(n)$. Hence, one defines the covariant derivative:
$$(DA)_a = dA_a + \omega_a^{~b}A_b~,$$
where $\omega_a^{~b} = \left(\omega_a^{~b}\right)_{\mu}~dx^{\mu}$ is a one-form under coordinate transformations. Demanding that $DA'_a = M_{a}^{~b}\left(DA_b\right)$ then gives us that $\omega_a^{~b}$ must transform inhomogenously under $O(n)$. So far is clear to me.
However, if we consider the covariant derivative of the dual frame $e^a$, this is defined to be:
$$De^a = de^a + \omega^a_{~b}\wedge e^b $$
Why does the wedge product become important here but not in the previous case, since $A_a$ is also the projection of a one-form in our frame?