I believe the first equation should be $\alpha_2r + 2a = \alpha_1r$
In the diagram you have shown, the top pulley (pulley 1) is fixed and can only rotate, while the bottom one (pulley 2) can move.
Now, the length of the belt should remain constant, and it being a rough belt indicates that we can neglect slipping at the points of contact.
$\alpha_1r$ is the acceleration of the belt downwards at the point of contact with pulley 1.
When pulley 2 moves down by some distance x, it will cause the belt to move down by 2x. This is because the motion will require x length of the belt to be added to either side of the pulley, which is achieved via the whole belt moving by 2x.
The same logic applies to velocities and accelerations.
So, acceleration of belt downwards at point of contact with pulley 2 is $2a$.
But this is only due to the translational motion.
If the pulley is only rotating, it would cause the belt to accelerate by $\alpha_2r$ to avoid slipping.
Now, we can superpose both motions. So the acceleration of belt =
$$\alpha_2r + 2a = \alpha_1r$$
The second equation comes from the no-slipping condition at pulley 2.
To be more specific, look at the rightmost point of contact. The belt there is stationary, so pulley 2 is undergoing pure rolling.