Transforming to Riemann Normal Coordinates I'm trying to do an ostensibly straight forward exercise in Zee's GR book, but am failing to understand the basic steps involved. Here's the suggestion on page 89:

"Suppose you were given a space described by the metric $ds^2 = dr^2 + r^2d\theta^2$ [...] Calculate the curvature by first transforming polar coordinates into locally flat coordinates at the point $(r,\theta) = (r_\star,0)$ [...] Then extract the combination of the $B_{\mu \nu ,\lambda\sigma}$s giving the intrinsic curvature [where $B_{\mu \nu,\lambda \sigma }$ is simply the second order expansion of $g_{\mu\nu}$]."

I know that Riemann Normal Coordinates must satisfy $g_{\mu \nu}|_P = \eta_{\mu \nu}$ and $\Gamma^\mu_{\nu\lambda}|_P = 0$, but it's not obvious to me how to satisfy this. My gut tells me I should create some ansatz for the new metric $g'_{\mu\nu} = \frac{\partial x^{\prime \mu}}{\partial x^\lambda}\frac{\partial x^{\prime \nu}}{\partial x^\sigma}g_{\lambda\sigma}$ and try to finagle my way to a transformation which satisfies the above, but that can't be the most efficient way of doing this. Sorry if this is obvious; it's my first time attempting to self-teach. If you need more context, I'd be happy to provide it. 
 A: A. Zee suggests we should go through all the steps described on page 87 - 88. I just finished the calculation, which looks quite cumbersome. 
Step 1. We first diagonalize the metric at the point $(r,\theta) = (r_*,0)$, $ds^2=dr^2+r_*^2d\theta^2$. Since it is already diagonal, we need no more change of coordinates.
Step 2. Scale each coordinate by an appropriate factor so that the diagonal elements become 1. Apply the change of coordinates, $r=r',\theta = \theta'/r_*$, so the metric in new set of coordinates is $$ds^2=dr'^2+\biggl(\frac{r}{r_*}\biggr)^2d\theta'^2=dr'^2+\biggl(1+2\frac{r-r_*}{r_*}+\biggl(\frac{r-r_*}{r_*}\biggr)^2\biggr)d\theta'^2. (*)$$
In the second equality, we expand $(r/r_*)^2$ in terms of $r-r_*$.
Step 3. Apply the change of coordinates again to get rid of the linear term 
$$2\frac{r-r_*}{r_*}$$
in the metric.
\begin{align}
r' = r''+ \frac{a_{111}}{2}(r''-r_*)^2+a_{112}(r''-r_*)\theta''+\frac{a_{122}}{2}\theta''^2\\
\theta'=\theta''+\frac{a_{211}}{2}(r''-r_*)^2+a_{212}(r''-r_*)\theta''+\frac{a_{222}}{2}\theta''^2
\end{align}
where the $a$'s are coefficients to be determined. Then
\begin{align}
dr' = dr''[1+a_{111}(r''-r_*)+a_{112}\theta''] + d\theta''[a_{112}(r''-r_*)+a_{122}\theta'']\\
d\theta' = dr''[a_{211}(r''-r_*)+a_{212}\theta'']+d\theta''[1+a_{212}(r''-r_*)+a_{222}\theta'']
\end{align}
Substitute it into eq. (*) and only retain first order of $(r-r_*)$ and $\theta''$ we get
\begin{align}
ds^2 &= dr''^2[1+2a_{111}(r''-r_*)+2a_{112}\theta'']\\
     &+d\theta''^2[1+2(r''-r_*)(\frac{1}{r_*}+a_{212})+2a_{222}\theta'']\\
     &+d\theta''dr''[(a_{112}+a_{211})(r''-r_*)+(a_{122}+a_{212})\theta'']
\end{align}
We want the linear terms vanish, so we get
\begin{align}
a_{111}=0,a_{112}=0,a_{122}=\frac{1}{r_*}\\
a_{211}=0,a_{212}=-\frac{1}{r_*},a_{222}=0
\end{align}
Now we know that
\begin{align}
dr' = dr''+\frac{1}{r_*}\theta'' d\theta''\\
d\theta' = -\frac{1}{r_*}\theta''dr''+\biggl[1-\frac{r''-r_*}{r_*}\biggr]d\theta''\\
r = r' = r''+\frac{1}{2r_*}\theta''^2
\end{align}
Now plug back to eq. (*) and retain first and second order of $(r''-r)$ and $\theta''$, we finally have
\begin{align}
ds^2 = \biggl[1+\frac{1}{r_*^2}\theta''^2\biggr]dr''^2 -\frac{2}{r_*^2}\theta''(r''-r_*)d\theta''dr''+\biggl[1+\frac{2}{r_*^2}\theta''^2-\frac{2}{r_*^2}(r''-r_*)^2\biggr]d\theta''^2
\end{align}
Notice in new coordinates the metric has quadratic expansion form. The desired $B_{\mu\nu,\lambda\sigma}$ is the coefficients in the above expression,
\begin{align}
B_{12,12}=-\frac{1}{2r_*^2},B_{22,11}=-\frac{2}{r_*^2},B_{11,22}=\frac{1}{r_*^2}
\end{align}
and the intrinsic curvature is $2B_{12,12}-B_{22,11}-B_{11,22}=0$ as expected because the given metric is just polar coordinates.
A: Tony Zee is probably refering to the fact that in locally geodesic coordinates centered at $x^\mu=0$ we have
$$
g_{\mu\nu}(x)= \delta_{\mu\nu}- \frac 13 R_{\mu\sigma \nu\tau}(0) x^\sigma x^\tau + O(|x|^3),
$$
and 
$$
{\Gamma^{\lambda}}_{\mu\nu}(x)= -\frac 13 (R_{\lambda\nu\mu\tau}(0)+R_{\lambda\mu\nu\tau}(0))x^\tau+ O(|x|^2).
$$
Thus knowing the metric to second order allows us to read off the curvature tensor.
Similarly we we can construct  local vielbein frames in which we have
$$
e^{*a}_\mu(x)= \delta_{a \mu}- \frac 16  R_{a \sigma \mu\tau}(0) x^\sigma x^\tau +O(x^2)
$$ 
and the corresponding frame-bundle connection coefficients are
$$
{\omega^a}_{b\mu}(x)=- \frac 12 {R^a}_{b\mu\tau}(0)x^\tau+O(|x|^2).
$$
