Wikipedia gives the metric as:
I found the equation below on line, but have not been able to find it again.
K = Gaussian curvature.
Because the circumference equation involves, K, it could be derived from the definition of Gaussian curvature near a pole shown below, not from integrating ds:
In which higher exponential terms of R go to zero rapidly. R is defined as the geodesic radius, traveling over the pole.
I am not surprised that you cannot find this formula for $C(r)$ again because it seems wrong.
To start, we should not confuse the polar coordinate $r$ with the geodesic distance $r$. Best to forget $ds^2=dr^2+r^2\,d\theta^2+r^2\sin^2d\phi^2$ (using $r$ as a polar coordinate) completely as it is not needed. The unit sphere has radius one.
The geodesic distance from the north pole $(1,0,0)$ of a point on the unit sphere that is at $(1,\theta,0)$ is $$ r=\theta. $$ The circumference $C(r)$ of this geodesic circle with radius $r$ is $C(r)=2\pi\sin\theta=2\pi\sin r\,.$ Using the well known power series of $\sin$, $$\tag{1} C(r)=2\pi\Bigg\{r-\frac{r^3}{3!}+\frac{r^5}{5!}-\frac{r^7}{7!}+...\Bigg\}\,. $$ By the Bertrand-Diquet-Puiseux theorem the Gaussian curvature is $$ K=\lim_{r\downarrow 0}3\frac{2\pi r-C(r)}{\pi r^3}\,. $$ Using (1) this is $$ K=\lim_{r\downarrow 0}\frac{6}{r^3}\Bigg\{\frac{r^3}{3!}-\frac{r^5}{5!}+\frac{r^7}{7!}+...\Bigg\}=1 $$ as expected. You could obviously write (1) as $$\tag{1} C(r)=2\pi\Bigg\{r-\frac{Kr^3}{3!}+\frac{Kr^5}{5!}-\frac{Kr^7}{7!}+...\Bigg\} $$ but that's a fairly useless improvement for the unit sphere.