general relativity - Geodesic Deviation between Test Particles from Gravitational Wave

I'm having trouble understanding how Carroll (Spacetime and Geometry, p.296) explains the effect of a passing gravitational wave on test particles.

If we have two geodesics with tangents $\vec{U}$, $\vec{U'}$ that begin parallel and near each other, and $\vec{S}$ is a vector connecting one geodesic to another at equal affine parameter values, then the equation of geodesic deviation is: $$\begin{align*} \frac{D^2}{d\tau^2}S^\mu = R_{\ \ \nu\rho\sigma}^\mu U^\nu U^\rho S^\sigma. \tag{7.103} \end{align*}$$ We work in the weak-field limit and the transverse-traceless gauge. If we assume our particles on the geodesics are moving slowly, then $$\vec{U} \approx (1,0,0,0),\tag{7.104}$$ so: $$\begin{align*} \frac{D^2}{d\tau^2}S^\mu = R_{\ \ 00\sigma}^\mu S^\sigma.\tag{*} \end{align*}$$ Now the bit I don't understand is how Carroll is able to turn the double covariant derivative on the left into a simple double-derivative with respect to $t$: $$\begin{align*} \frac{\partial^2}{\partial t^2}S^\mu = R_{\ \ 00\sigma}^\mu S^\sigma.\tag{**} \end{align*}$$ Carroll's reasoning is that "for our slowly moving particles we have $\tau = x^0 = t$ to lowest order", but I don't know what he means. I just don't understand why the Christoffel symbols vanish in the covariant derivatives. I have read several books about this. Some say the Christoffel symbols vanish because we work in a local inertial frame. But then why doesn't the Riemann tensor on the RHS also vanish?