The vielbein postulate says that
$$\nabla_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu-\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma=0.$$
$\nabla$ is the coordinate covariant derivative, and $\Gamma^\sigma_{\mu\nu}$ is connection defined as $$\Gamma^\sigma_{\mu\nu}=\tilde{\Gamma}^\sigma_{\mu\nu}+K^\sigma_{\mu\nu},$$
where $\tilde{\Gamma}^\sigma_{\mu\nu}$ is the Christoffel symbol and $K^\sigma_{\mu\nu}$ is the contorsion.
The Lorentz covariant derivative however is defined as $$D_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu,$$ and the vielbein postulate then implies that $$D_\mu e_v^{\,a}=\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma.$$
Now, I want to take the Lorentz covariant deriavtive of the vielbein determinant, namely $$D_\mu\left(\sqrt{|g|}\right),$$ which we can take in the local coordinates to be $$D_\mu\left(\sqrt{-\text{det}(e_\alpha^{\,\,a}e_\beta^{\,\,b}\eta_{ab})}\right)=D_\mu\sqrt{e^2}=D_\mu|e|.$$
I am stuck here though as I'm unsure how one exactly takes this derivative. I would assume it should become a partial derivative as this quantity is just a scalar. Also, if one goes back a few steps and considers the relationship between the Christoffel symbols and the metric, we could just write this as
$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\,\tilde{\Gamma}^\sigma_{\sigma\mu}.$$
Is this correct, or is there a better way to express this derivative?
EDIT:
Just in response to the answer given by @Immanuel (to avoid a long comment section), I will try to make clear my confusion with their answer.
The Lorentz covariant derivative is $$D_\mu V^a=\partial_\mu V^a+\omega_{\mu\,\,b}^{\,\,a} V^b,$$ where the spin connection $\omega$ looks like $$\omega_{\mu\,\,b}^{\,\,a}=\tilde{\omega}_{\mu\,\,b}^{\,\,a}+K_{\mu\,\,b}^{\,\,a}.$$
So I suppose my first point of confusion is why you have a $\Gamma$ instead of an $\omega$ in your definition of the Lorentz covariant derivative of a density.
If my definition is correct, shouldn't the Lorentz covariant derivative of a density be
$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|},$$ which becomes $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^{\lambda}+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|}.$$
If $\omega_{\mu\,\,a}^{\,\,a}=0$, then we would have $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^\lambda,$$
so I hope that clears up some of my confusion.