Extending Maxwell's Equations from Flat Spacetime To Curved Spacetime Assume we are working on a Minkowski (i.e. flat) spacetime. 
Let $A^{\mu} = ( \phi/c, \textbf{A})$ be the contravariant potential four-vector. Then, assuming a covariant Minkowski metric of $\eta_{\mu \nu} = \textrm{diag}[+, -, -, -]$, we have that $A_{\mu} = ( \phi/c, -\textbf{A})$ is the covariant potential-four vector.
We also have that $\alpha = A_{\mu} dx^{\mu}$ is the potential one-form.
We then define $ F = d\alpha = \frac{1}{2} (\partial_\mu A_\nu - \partial_\nu A_\mu) dx^\mu \wedge dx^\nu$ to be the electromagnetic two-form.
Now, let $J^{\mu} = (c\rho, \textbf{J})$ be the contravariant current four-vector. 
Then, $ J = \frac{1}{6} J^\mu \epsilon_{\mu \alpha \beta \gamma}dx^\alpha \wedge dx^\beta \wedge dx^\gamma$ is the current three-form.
With these definitions, Maxwell's equations become
\begin{equation}
dF = 0 \; \; \; (\textrm{i})
\end{equation}
\begin{equation}
d(*F) = J \; \; \; (\textrm{ii})
\end{equation}
(Recall that $*$ is the Hodge Star operator).
Now, to extend these Maxwell equations to a curved spacetime, it appears that we must alter the current three-form:
\begin{equation}
J = \frac{1}{6} \sqrt{|g|} J^\mu \epsilon_{\mu \alpha \beta \gamma}dx^\alpha \wedge dx^\beta \wedge dx^\gamma \; \; \; (\textrm{iii})
\end{equation}
Here, $\sqrt{|g|}$ is the square root of the absolute value of the determinant of the covariant metric on the Riemannian manifold were are working with.
With this new definition of $J$, Maxwell's equations are just equations (i) and (ii).
My question is the following. Why does simply modifying the current three-form to include the "natural" pseudo-Riemannian volume form $\sqrt{|g|} dx^\alpha \wedge dx^\beta \wedge dx^\gamma$ allow us to use the flat spacetime formulation of Maxwell's equations in curved spacetime? 
 A: Per se, this doesn't have anything to do with relativity or curvature: The factor of $\sqrt{|g|}$ comes in for a similar reason that has the determinant of the Jacobian pops up in the substitution formula for integration of multiple variables: When integrating, you need to account for the volume of the unit cell spanned by your coordinate frame. So if you're using generic curvilinear coordinates instead of pseudo-Euclidean ones, you need to add it to the expression for Minkowski spacetime as well.
A: The idea of generalizing laws to curved space-time is to notice that we actually live in a curved space-time ourselves. What we know as "flat space-time equations" are, in fact, equations in curved space-time derived/discovered in our local (almost-)inertial frame. We can then derive their curvilinear form by simply transforming to a general frame. This is done mostly by replacing any use of the Minkowski metric structure by a general pseudo-Riemannian one. 
Specifically in the case of Maxwell equations the differential-geometry form is almost covariant already. But notice that you are using a metric structure at two points, and both can be characterized as using the Hodge dual. I use a definition of a Hodge dual that takes me from $\Lambda^{k} T^*\! \mathcal{M}$ to $\Lambda^{(n-k)}T^{*}\!\mathcal{M}$, where $n$ is the dimension of the manifold $\mathcal{M}$ (this is unlike the definition used in the wikipedia page). The most practical way to define this Hodge dual for any form $\alpha \in \Lambda^{k} T^*\! \mathcal{M}$ is to require that
$$(*\alpha) \wedge \beta = \beta(\alpha^{\#k}) \omega, \forall \beta \in \Lambda^{k} T^*\! \mathcal{M},$$ 
where $\alpha^{\#k}\in T^k\! \mathcal{M}$ is obtained by raising all the indices of $\alpha$, and $\omega$ is the pseudo-Riemannian volume form $\omega = \sqrt{|g|} \mathrm{d}x^1\wedge...\wedge \mathrm{d}x^n$ (note that $\sqrt{|\eta|} = 1$ in Cartesian/Minkowski coordinates and we specialize to $n=4$). Now you can see that the Hodge dual can be obtained by contracting the volume form $\omega$ with $\alpha^{\#k}$.
Back to the Maxwell equations. What you call the current 3-form is in fact the Hodge dual of the current 1-form $j = j_\mu \mathrm{d}x^\mu, J = *j $. In your statement you use the generation of the dual by contracting with the volume form, which would usually be stated as
 $$J \equiv *j = \omega (j^{\#},\cdot,\cdot,\cdot) = \iota_{j^{\#}}\omega = \frac{1}{3!} j_\mu g^{\mu\nu} \sqrt{|g|}\epsilon_{\nu\lambda\kappa\gamma} \mathrm{d}x^\lambda\wedge\mathrm{d}x^\kappa\wedge\mathrm{d}x^\gamma$$
Here you can identify $j^\# \equiv j_\mu g^{\mu\nu} \partial_\nu = J^\nu \partial_\nu$ as your current vector (but beware that on metric manifolds objects with raised and lowered indices are considered as identical objects expressed in a different way).
In summary, the covariant statement of the Maxwell equations is
$$\mathrm{d}F = 0\,,$$
$$\mathrm{d}(*F) = *j\,,$$
where you have to remember that the Hodge dual is now generated by the general metric $g$. The last line is actually very often written as $*[\mathrm{d}(*F)] = j$ (which is equivalent to the one above since the Hodge star is a dual).
