Let's say I have a metric for spacetime that looks like this: $$ ds^2 = -c^2dt^2+a^2(t) \left[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2\right] $$ As I understand it, this is the Reduced Circumference version of the formula where we basically take the circumference as it's measured, but then reduce the radius such that $C=2\pi D$ where $D$ is the reduced radius. In that way, basic trig operations can still return usable results.
I also know from basic trigonometry on a sphere that $$D=\frac{\sin\left(r\sqrt{k}\right)}{\sqrt{k}}$$ when the curvature is positive. Conceptually, this appears to be the same thing as the Reduced-circumference polar coordinates. Is there a way to derive this relation from the metric? If so, please show the steps.
I am basically trying to find the relation between $r$ and $\theta$ in a space with positive curvature, $k$ using my metric for that space.
First part
It is customary to use capital letter $K$ to denote Gaussian curvature ( not $k$). For positive Gauss curvature let $ K= 1/a^2$
I am supposing that you are trying to find a relation between $D$ and $r$.
On the question how to exploit the metric in order to find shrunk or reduced circumferential distance $r$ from arc distance $D$ of 2D spherical surface.
For a given latitude or longitude ( same $(\phi, \theta)$ so $ d\phi=0, d\theta=0) $ these two terms vanish so we do not consider parallel circle, operate on cylindrical mode to obtain formula for geodesic distance $D$ between a pole and a geodesic parallel:
$$ dD^2= \dfrac{dr^2}{1-Kr^2}=\dfrac{dr^2}{1-r^2/a^2} \; (r< a) $$
$$dD =\dfrac{dr}{\sqrt{1-r^2/a^2}} $$
$$ D =\int \dfrac{dr}{\sqrt{1-r^2/a^2}}= a \sin^{-1}\dfrac{r}{a}$$
so that the direct Reduced Circumferentially determined radius
$$ r = a \sin \dfrac{D}{a} = a \sin D \sqrt K \; (r<a) \tag1 $$
From trigonometry of a sphere we have
$$ \dfrac{r}{a} =\sin \dfrac{D}{a}= \sin \phi_{co-latitude}\quad r<a \tag2 $$
but not
$$ \dfrac{D}{a} =\sin \dfrac{r}{a} \tag3$$ which as an error source in your simple trigonometric calculation is clear.. as $ r$ cannot be greater then $D$.
Also reduced circumference is
$$ 2 \pi r < 2 \pi D $$
Second part
You are in other words looking at geodesic polar coordinates for the general case with North pole as center for these coordinates. Relation between $D$ and $\theta$ in general involves metric. Even if $D$ is an uncomfortable symbol,lengths are according to the metric:
$$ ds^2= dD^2+G_{D,\theta} \;d\theta^2 $$
involving first fundamental form coefficient $G_{D,\theta} $.
It should be noted for generality that for $K<0$ circumferential dimension increases in hyperbolic geometry.