A smaller sphere purely rolling down without slipping over another larger sphere 
In this question we have to find out the angular velocity of the smaller sphere about its own axis at the instant it leaves the surface of the larger sphere and it is given that the smaller sphere is purely rolling over the larger sphere throughout its motion.
I tried it using two methods.
Method 1- 
The angle at which the sphere leaves the larger sphere can be found out using
$$mgcos(\theta) = mv^2/(R+r)$$
Next applying energy conservation about the smaller sphere's own axis
$$\frac{1}{2}(mv^2) + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mg(R+r)(1-cos(\theta))$$
The first term in this equation denotes the translational kinetic energy and the second term the rotational kinetic energy.
from here we get value of v and then $angular velocity= \frac{v}{r}$.
Method 2-
using the 1st condition as in method 1 we get, angle at which the sphere leaves the larger sphere
Next I tried to apply the energy conservation equation about the axis of the larger sphere, and am ending up with a wrong answer. 
On the left hand side the first term is the rotational kinetic energy of the smaller sphere about the centre of the larger sphere. The second term denotes the rotational kinetic energy of the smaller sphere(due to rotation about its own axis)
$$\frac{1}{2}(\frac{2}{5}mr^2 + m(R+r)^2)(\frac{v}{R+r})^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mg(R+r)(1-cos(\theta))$$
Please note that it is a genuine doubt encountered while trying to do the question through other methods.
 A: Which is right?
So let's first establish what is correct. Suppose we have a circle of radius $r$ and mass $m$ centered on a circle of radius $\bar R$ (not quite what you have defined; in your case $\bar R = R + r$). We'll say that the position of the $r$-circle on the $\bar R$-circle is angle $\phi$ and, from the angle $\phi$, we will measure an angle $\theta$ around the smaller circle.
The position of a little piece of mass on the $r$-circle is therefore, in xy-coordinates, $\bar R [\cos\phi, \sin\phi] + r [\cos(\phi+\theta), \sin(\phi+\theta)]$ and its velocity is $$\vec v(\phi,\theta) = \bar R ~\Omega [-\sin\phi,\cos\phi] + r(\omega + \Omega) [-\sin(\phi+\theta), \cos(\phi+\theta)],$$ where I have defined $\Omega = d\phi/dt$ and $\omega = d\theta/dt.$
Its squared velocity is therefore:$$v^2 = (\bar R ~\Omega)^2 + [r(\omega + \Omega)]^2 + 2\bar R ~\Omega~r(\omega + \Omega)\cos(\theta).$$
Integrating with a linear mass density $dm = \lambda~r~d\theta,$ $m = 2 \pi r~\lambda,$ the big effect is that the cosine disappears and we're just left with:$$K = \int dK = \int \frac12 ~dm~ v^2 = \frac12 m \left[\bar R^2 \Omega^2 + r^2 (\omega+\Omega)^2\right]$$
Now for rolling without slipping on a stationary surface we need a velocity of zero at the angle $\theta = \pi,$ so $\bar R\Omega - r (\omega + \Omega) = 0,$ and $\omega = \frac{\bar R - r}{r} ~\Omega.$
So in terms of $\Omega$ (which for you is $v/(R + r)$) and $R = \bar R - r,$ and generalizing from our $mr^2$ to a moment of inertia $I$, we can see that the proper expression is $$K = \frac12 \left[m (R+r)^2 + I~\left(\frac Rr + 1\right)^2\right]~\Omega^2 = \frac12 m v^2 + \frac12 I \left(\frac vr\right)^2.$$So your first method is correct when we carefully analyze it from the second method's perspective.
What went wrong?
Now that we know what's right, what went wrong in your second argument?
Your second method appears to suffer from an unnecessary duplication of $I$ and the assumption that $\omega = v/r$ rather than the correct $\omega = R v / [r (R + r)].$ This difference probably comes from you forgetting that your new coordinates are rotating with respect to the old coordinates, so some of the $v/r$ rotation is already incorporated in them. This presumably generates the discrepancy factor when fully taken into account.
