Exterior derivative of 2-forms and divergence I'm working through A Geometric Approach to Differential Forms. The deriative of a $2$-form $\omega$ in $\mathbb{R}^3$, denoted $d\omega$ and operating on vectors $U, V, W \in T_p \mathbb{R}^3$, is defined as
$$d\omega(U, V, W) = \nabla_U \omega(V, W) - \nabla_V \omega(U, W) + \nabla_W \omega(U, V)$$
where $\nabla_U \omega(V, W)$ denotes the directional derivative of $\omega(V, W)$ in the direction $U$.
Suppose that we have a $2$-form $\omega = f(x, y, z) \ dx \wedge dy + g(x, y, z) \ dy \wedge dz + h(x, y, z) \ dx \wedge dz$. One way to calculate $d\omega$ is to realize that it must have the form $d\omega = a(x, y, z) \ dx \wedge dy \wedge dz$. Geometrically, we can think of $d\omega$ as taking three vectors, calculating the signed volume of the parallelogram formed by those three vectors, and then applying a scaling factor $a(x, y, z)$. So one way to determine $a(x, y, z)$ is to use the above definition of $d\omega(U, V, W)$ to see how it acts on vectors corresponding to a parallelogram of signed volume $1$.
If I take $U = (1, 0, 0)$, $V = (0, 1, 0)$, and $W = (0, 0, 1)$, and I go through the computations, I arrive at
$$d\omega = \left( \frac{\partial g}{\partial x} - \frac{\partial h}{\partial y} + \frac{\partial f}{\partial z} \right) \ dx \wedge dy \wedge dz.$$
This seems to be right as far as I can tell. In particular, the second term is given by $\nabla_V \omega(U, W) = \partial h / \partial y$, with the negative sign coming from the alternating signs in the definition of $d\omega(U, V, W)$. Explicitly, my calculation is
$$\nabla_V \omega(U, W) = \nabla \left( \underbrace{f \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}}_0 + \underbrace{g \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}}_0 + \underbrace{h \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}}_h \right) \cdot V = \frac{\partial h}{\partial y}.$$
Furthermore, I know that there's a connection between the exterior derivative of a $2$-form in $\mathbb{R}^3$ and the operation of the divergence. From, e.g., the Wikipedia description, it seems that divergence is
$$(g, h, f) \mapsto \frac{\partial g}{\partial x} + \frac{\partial h}{\partial y} + \frac{\partial f}{\partial z}$$
where I've ordered $f, g, h$ for consistency with the representation of $\omega$.
The difference here compared to my expression for $d\omega$ is that the term $\partial h / \partial y$ is positive rather than negative.
Whence the difference?

You are using $dx\wedge dz$ and the wiki page is using $dz\wedge dx$. So there is no difference. 
remark: In general for an $n-1$ form, one usually insert some $(-1)$'s to deal with this: a general $n-1$ form $\alpha$ is written as 
$$\alpha = \sum_{i=1}^n (-1)^{i-1} \alpha_i dx^1 \wedge \cdots \wedge\widehat{dx^i} \wedge\cdots \wedge dx^n$$
so that 
$$d\alpha = \left( \frac{\partial \alpha_1}{\partial x^1}+ \cdots +  \frac{\partial \alpha_n}{\partial x^n}\right) dx^1 \wedge \cdots \wedge dx^n. $$
