Conformal transformation of the divergence

Here's an relatively straightfoward, coordinate-free way to approach the problem:

The divergence of a metric $g$ actually depends only on its volume form $\mu$ (for either orientation, it doesn't matter which). It can be characterized by $$\mathcal{L}_X \mu = (\text{div } X) \mu.$$

Now, the volume form $\widehat{\mu}$ of a conformally related metric $\widehat{g} = e^{2f} g$ is $$\widehat{\mu} = e^{n f} \mu;$$ substituting in the above characterization of the divergence---and using the Leibniz identity $$\mathcal L_X (h \alpha) = (X \cdot h) \alpha + h \mathcal L_X \alpha$$ for a function $h$ and form $\alpha$---gives an identity relating the divergences of the two metrics.