$P_0-P_1 :$
With your chosen cartesian coordinate, consider the cue ball with radius $r$ start with initial velocity $\vec{v} = (0,v_0,0)$ and initial angular velocity $\dot{\theta} = (\omega^*,0,0)$ (so its back-spin). I assume we know the distance from $P_0$ to $P_1$ before its hit another ball. Call the distance $s$. Because its spin backward, the friction force must act on the ball to the negative direction of its velocity so the total force will be $\vec{F} = (0,-f,N-mg)$, where $f$ is its kinetic friction with magnitude $f=\mu_kN$. By Newton's law in the $y-$axis and torque equation
$$
-f = m a_y \quad \text{and} \quad -fr = I \alpha
$$
At this situation $a_y \neq \alpha r$ because its slipping. With distance $s$ the final velocity will be
$$
v^2 = v_0^2 +2a_y s = v_o^2 -2\frac{f}{m}s = v_0^2 - 2\mu_kgs, \quad v = \sqrt{v_0^2 - 2\mu_kgs} \qquad (1)
$$
To compute the final angular velocity we need the time travel of the ball with distance $s$, that is
$$
s = v_ot + \frac{1}{2} a_y t^2 = v_0t - \frac{1}{2}\mu_kg t^2 \iff \frac{1}{2}\mu_kg t^2 -v_0t + s = 0
$$
and solve for $t$. The angular velocity will be
$$
\omega = \omega^* + \alpha t = \omega^*- \frac{fr}{I} t = \omega^* - \frac{\mu_k mgr}{I} t \qquad (2)
$$
So at $P_1$ the velocity is $\vec{v} = (0,v,0)$ and $\dot{\theta} = (\omega,0,0)$ and it still slipping (otherwise the ball will not move back after hit the another ball as shown by the video).
$P_1 - P_1 $ (collision) :
Consider the collision to be elastic (no deformation of the balls) so the kinetic energy conserved
$$
\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I \omega'^2 + \frac{1}{2} M V^2 + \frac{1}{2}I_M \omega_M^2\qquad (3)
$$
with $v'$ and $\omega'$ are velocity and angular velocity of the cue ball whereas $M,I_M, V, \omega_M$ are the mass, moment of inertia, velocity and angular velocity of the target ball respectively. Now consider the Impulse momentum equation from Newton's law below
$$
\sum \vec{F}_i = \sum \frac{d\vec{p_i}}{dt} \quad \text{or} \quad \sum \vec{F_1} + \sum \vec{F_2} = \frac{d\vec{p_1}}{dt} + \frac{d\vec{p_2}}{dt}
$$
integrate them, we have impulse momentum equation
$$
(\vec{p_1})_{\text{initial}} + (\vec{p_2})_{\text{initial}} + \int (\sum \vec{F_1} + \sum \vec{F_2}) dt = (\vec{p_1})_{\text{final}} + (\vec{p_2})_{\text{final}}
$$
Where index $i=1,2$ represent out system of two ball, and the sum above represent all force terms that act on each balls. Those force are the internal forces : contact force and frictions, and the external forces : weight, normal forces and frictions from the table. The internal forces are comes in pairs so if we add the forces $\sum \vec{F_1} + \sum \vec{F_2}$, those will cancels each other. The normal force and the weight will cancel each other. So we just have frictions from the table that contribute, and in fact in the small interval (as they collide) the product of theese constant force and the time interval is so small so they can be neglected,
$$
\int (\sum \vec{F_1} + \sum \vec{F_2}) dt = \int_{0}^{\delta} (\vec{f_1} +\vec{f_2} ) dt = (\vec{f_1} +\vec{f_2}) \delta \approx 0
$$
Therefore the momentum is conserved. So
$$
(\vec{p_1})_{\text{initial}} + (\vec{p_2})_{\text{initial}} = (\vec{p_1})_{\text{final}} + (\vec{p_2})_{\text{final}} \qquad (4)
$$
We know that the target ball is not move at the beginning so $(\vec{p_2})_{\text{initial}} = \vec{0}$. Also the direction of the velocity of the target ball will be perpendicular to the tag line so $\vec{p_2})_{\text{final}} = M\vec{V} = (-MV \cos \alpha, MV \sin \alpha,0)$. And also $(\vec{p_1})_{\text{initial}} = (0,mv,0)$. We dont know the direction of the cue ball after collision so lets just say $(\vec{p_1})_{\text{final}} = (mv'_x,mv_y',0)$. Therefore $x$-component and the $y-$component of the momentum eq. $(4)$
$$
0 = mv_x'- MV \cos \alpha, \qquad mv = mv_y' + MV \sin \alpha \qquad (5)
$$
rewrite and squared both sides
$$
(mv_x')^2 = (MV \cos \alpha)^2, \qquad (mv_y')^2 = (mv - MV \sin \alpha)^2
$$
and then add them, you'll have
$$
m^2v'^2 = m^2v^2 + M^2 V^2 - 2mM vV \sin \alpha \qquad (6)
$$
Now we have to make other assumption that the angular velocity of the ball does not affected by the collision so in the equation $(3)$, $\omega'=\omega$ and $\omega_M = 0$ (actually i think this is the case in pool because the surface of the balls is very smooth). Otherwise it will be very difficult to solve because now the angular velocity not vanish due to the horizontal friction and vertical friction (i'm sorry i dont have the picture) and we have to analyze them using another equation that is angular impulse-momentum equation derived from integrating torque equation. By this assumption $(3)$ become
$$
mv^2 = mv'^2 + MV^2
$$
Combine this and $(6)$ we have
$$
0 = V^2 (1+ \frac{M}{m}) - 2vV \sin \alpha \implies V = \frac{2v \sin \alpha}{1+ \frac{M}{m}}
$$
And therefore by $(5)$ you'll have $v' =\sqrt{ v_x'^2+v_y'^2}$ also.
$P_1 - P_2$ (after collision) :
We have to analize two ball after the collision. Lets consider the easier first that is the target ball. Right after the collision, the target ball moving with velocity $V$ in the direction of the line perpedicular to the tag line. Call this line as $y'-$axis and the tag line $x'-$axis. Because of this, the target ball must slip for a moment and we have to find when this slip is over. By Newton's Law and torque eq. (along $y'$-axis) and the fact that $f = \mu_kMg$
$$
-f =-\mu_k Mg= Ma, \qquad -fr=-\mu_kMg r=I_M \alpha
$$
So the velocity and the angular velocity of this target ball will be
$$
V_M = V + at = V-\mu_k g t, \quad \omega_M = \omega_M0 + \alpha t = 0 - \frac{\mu_kMgr}{I_M} t
$$
The minus sign in the $\omega_M$ just means that the roatation is top spin. The non-slip condition happen (at some time $t^*$) if the magnitude of $V_M$ and $\omega_M r$ are equall. So set $|V_M| = |\omega_M|r$
$$
V - \mu_k g t^* = \frac{\mu_kMgr}{I_M} t^* \implies t^* = \frac{V}{\mu_k g(1+ \frac{rM}{I_M})}
$$
Put this back to the $V_M = V + at^* = V-\mu_k g t^*$ we have the final velocity of the target ball until non-slip condition
$$
V_M = \frac{V}{\frac{I_M}{Mr}+1}
$$
And we're done with the target ball.
For the cue ball after collision, the cue ball has condition $\vec{v'} = (v_x',v_y',0)$ with back spin $\dot{\theta}= (\omega,0,0)$ as before collision. Because of the back spin, there is a friction force $\vec{f_1} = (0,-f_1,0)$ and because of the movement in the $+x$-direction (by $v_x'$) we have the second friction force $\vec{f_2}= (-f_2,0,0)$ with their resultant $\vec{f}=\vec{f_1}+\vec{f_2}$ satisfy $|\vec{f}| = \mu_kmg$. And to this point i simply have a doubt about how to proceed to analized this. I'll update it another time if i know.
