Given the Schwarzschild metric
$$ ds^{2} = -\Big(1-\frac{r_{s}}{r}\Big)c^{2}dt^{2} + \Big(1-\frac{r_{s}}{r}\Big)^{-1}dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}), $$
we can apply the approximation $(1-x)^{-1}\approx 1+x$ and get
$$ ds^{2} = -\Big(1-\frac{r_{s}}{r}\Big)c^{2}dt^{2} + \Big(1+\frac{r_{s}}{r}\Big)dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). $$
Taking $r_{s} = 2GM/c^{2}$ and $\Phi(r) = -GM/r$ gives
\begin{align}\tag{1} ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). \end{align}
This is nice and all. However, page 126 in Gravity by Hartle and eqn (12) in page 2 in this link give us something a little different:
$$ ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) (dx^{2} + dy^{2} + dz^{2}). $$
By changing to spherical coordinates, we get
\begin{align}\tag{2} ds^{2} = -\Big(1 + \frac{2\Phi(r)}{c^{2}}\Big)(c\, dt)^{2} + \Big(1 - \frac{2\Phi(r)}{c^{2}}\Big) dr^{2} + r^{2}\Big(1 - \frac{2\Phi(r)}{c^{2}}\Big)(d\theta^{2} + \sin^{2}\!\theta \, d\phi^{2}). \end{align}
Question: How do we get from (1) to (2)? In (2), why is there that factor in the angular part of the metric even though it is not there in (1)? Is there some substitution/rescaling I need to do (nothing obvious seems to work)? Or do we just say that (1) and (2) are merely approximately the same?
If this is a matter of approximation, then I am still confused, because I thought we can only ignore terms of order $O(1/c^{4})$. The difference between (1) and (2) is only of order $1/c^{2}$.
