Force on a solid cylinder that is rolling on an accelerating block 
Given: $m,R,I=mR^2/2, F,M$ and no traction between $B_2$ and the ground find $a_1, a_2$ (the accelerations of the CMs of $B_1$ and $B_2$ respectively).$B_1$ rolls on $B_2$ without sliding due to $T$.
  

 A: No, what would be the source of that force? there is no such force F', unless the problem explicitly put it as an external force.  T is responsible for  both the torque for rotation and the acceleration of the center of mass of the ball.
Given the specific moment of inetria the ball has, you get a set of apparently contradictory equations. The solution is that is consistent will be T=0, which means that this specific ball cannot roll without sliding.
UPDATE:
we can make the ball rolling if we apply a force F'.
Ib this case there are two solutions to the problem. You can choose to put the force F' in the same direction of F, or in the opposite one. In the first case, the solution is $F'=.5ma_1=T$. In this example both T and F' are in the same direction. In the second solution both F' and T have the same direction, but this time opposite to F. The acceleration is negative. and in this case $F'=T=.5 ma_1$
in the first case, the solution for a1 is: $F+T=Ma_2$, thus: $a_2=(F+F')/M$
In the second case you have: $F-T=Ma_2$, thus: $a_2=(F-F')/M$
I am not sure how you got $a_2 = \frac{F}{m+3M}$ 
A: The center of mass of an object accelerates as though all forces acting on that object act there; there is no need to invoke the "red force" in your diagram. The torque on the cylinder is given by $\Gamma = T\cdot R$ and of course the difference between $T$ and $F$ is the force that accelerates the lower block, e.g. $F-T = M\cdot a_2$; make sure that the sign conventions are what you want them to be (you draw one $T$ to the right and another to the left...)
A: UPDATE (Solution):
The center of mass of $B_1$ has a displacement of $$\vec x_1 = \vec x_{1,1} + \vec x_{1,2}$$ One due to RWS and the other due to its contact with the other body. Differentiation (twice) leads to $$\vec a_1 = \vec a_{1,1} + \vec a_{1,2} \xrightarrow{algebraic} a_1 = a_{1,2} - a_{1,1}$$. But $$T = m a_1$$ and $$\sum \tau = I \alpha_{\gamma(1)} \iff T = ma_{1,1}/2$$ Finally $$F-T = ma_2 = m a_{1,2}$$ combining the equations leads to $$a_1 = \frac{F}{m+3M}$$ and $$a_2 = \frac{3F}{m+3M}$$
