I am a bit confused by the physical meaning of radius vs radius of curvature, with regard to an ellipse.
For a standard ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
In this case, the $a$ and $b$ refer to the "radius of curvature" of the ellipse in the $x$ and $y$ direction respectively.
In contrast to the radius of curvature for an ellipse:
$$ \frac{(a^2 \sin^2t + b^2 \cos^2t)^\frac{3}{2}}{ab} $$
Let's say that at $t = 0$, we get a radius of curvature of $\frac{b^2}{a}$.
How does this value relate to the original equation, where I should get $x = b$ at $t = 0$ (since b is the "radius" in the x-direction) instead?
The radius of curvature $R$ of a curve at a point is the radius of the circular arc which ''best'' approximates the curve at that point. Here ''best'' means that the system given by the curve equation and the circle equation have a double root in the point of contact.
Properly $a$ is not a ''radius'' of the ellipse, but it is the radius of the circle with center $(0,0)$ and that pass in the vertex $(a,0)$ of the ellipse. This is not the circle that ''best approximates'' the ellipse in this point, because the system has also the solution $(-a,0)$.
The ''best'' circle is $$ \left[x-\left( a-\dfrac{b^2}{a}\right) \right]^2+y^2=\left(\dfrac{b^2}{a}\right)^2 $$ that has as radius the curvature radius $R=\dfrac{b^2}{a} $ as found in OP. And you can see that the system of this equation and the equation of the ellipse has double root in $(a,0)$.