For a disk within a ring, why is arclength proportional to $R*\theta$ but not $a*\phi$? Here is an example problem, why is $\phi$ not by itself in the constraint equation? Can someone explain this geometrically? It seems like the curve of the surface curves away from the disk, but that seems hand-wavy and incorrect. 

 A: Because the surface that the sphere is on isn't flat. If it were flat, the distance would just be $a\phi$. 
Think of it this way: Imagine a marking on the edge of the sphere that is originally facing upwards. On a flat track, after having rolled one round, $2\pi a$, the marking will still be pointing up. This is also true for a curved track. However, notice that it will no longer point toward the center of the curved track, which we will call the radial direction. In other words, the "upwards" direction no longer coincides with the radial direction, but is instead offset by an angle $\theta$.
In your case, which is a concave track, the sphere has to roll an angle $\theta$ less because it has translated by a distance $2\pi a$, but the track has curved towards it by an angle $\theta$. Which is to say, the upwards direction is now ahead of the radial direction by an angle $\theta$. Thus, it has to roll an angle $-\theta$ in order for the marking to point radially again. Hence the opposite signs of $\theta$ and $\phi$ in the equation.
For a convex track, the sphere has to roll an additional angle $\theta$ because it has translated by a distance $2\pi a$, but the track has curved away from it by an angle $\theta$. Which is to say, the upwards direction is now behind of the radial direction by an angle $\theta$. Thus, it has to roll an angle $\theta$ in order for the marking to point radially again.
Why do we bother with the radial direction at all? Well, this is because the center of the sphere and the point of contact with the track both lie on the radial direction. It is the only direction that doesn't change with respect to the track, always remaining perpendicular to it. Imagine a small circle travelling around a larger, stationary circle, with its point of contact fixed. By travelling one round, it would have rotated once, even though it never rolled with respect to the larger circle!
To summarize, no matter which direction you use, the center and the point of contact do not travel the same distance, which would be the case on a flat surface, and this discrepancy must be accounted for.
Check out the Coin Rotation Paradox for more details.
A: Just wanted to add a picture to better visualize the equation of constraint. Hope it helps someone out there. I've called the angle related to the rotation of the sphere $\gamma$ and I've drawn the situation for when $\gamma=\pi$.

A: 
\begin{align*}
&\text{Position vector }\\\\
 R_2&=
\begin{bmatrix}
  (R-a)\,\sin(\varphi_1)+a\,\sin(\varphi_2) \\
 (R-a)\,\cos(\varphi_1)+a\,\cos(\varphi_2) \\
\end{bmatrix}
\\\\
&\text{Velocity :}\quad v=\sqrt{\dot{R}_2^T\, \dot{R}_2}=v(\varphi_1\,,\varphi_2\,,\dot{\varphi}_1\,,\dot{\varphi}_2)
  \\
  &\text{At the contact point is: $\varphi_2=\varphi_1\,\Rightarrow\quad v=v(\dot{\varphi}_1\,,\dot{\varphi}_2) $}\\
  &\text{Motion without slipping}\quad \Rightarrow\quad 
  v_c=v(\dot{\varphi}_1\,,\dot{\varphi}_2)\overset{!}{=}0\\
  &\Rightarrow\\
  \dot{\varphi}_2&=-\frac{R-a}{a}\dot{\varphi}_1
  \quad \Rightarrow\quad  {\varphi}_2=-\frac{R-a}{a}{\varphi}_1
\end{align*}
