I must admit I cheated and looked at the solution, and the solution is really simple. I would probably have solved the equation of motion for the system, and it isn't necessary to go into that much detail.
That certainly isn't true during the compression phase because the wall to the left of $m_1$ is exerting a force on $m_1$ so there is an external force acting.
If you compress the spring a distance $x$ then the PE is $\tfrac12 kx^2$. At this moment both masses are at rest so the KE is zero.
When the mass $m_2$ has moved back to its initial position the PE in the spring is zero so the KE of $m_2$ is now $\tfrac12 k x^2$ and the mass $m_1$ is still at rest.
If the particle moved outwards from the centre of the disk at velocity $v_0$ then their relative velocity is going to be $v = r\omega$ where $r$ is the distance from the centre of the disk. $r = v_0t$.
So in the frame of the disk the momentum of the particle will be $p = mv = mv_0t$.