Here is a still from the video and in this case the term absolute in the context of the acceleration means magnitude relative to the ground.
![enter image description here](https://i.sstatic.net/Iy2Qh.jpg)
Applying Newton's second law in the ground frame of reference for each of the masses with up as positive:
$T-6g = 6(b+\frac g8)$ and $T - 7g = 7(\frac g8 -b)$
and with two equations you can solve for the two unknowns $T$ and $b$.
Notice that in the video up is positive fro the left hand mass and down is positive for the right hand mass which I prefer not to do.
The two equations are essentially the same because mutiplying my second equation by $-1$ produces the lecturer's second equation.
Now let me rearrange my two equations by putting the term which contains $\frac 98$ on the left hand side.
$T-6g-6\frac g8 = T-6(g+\frac g8) = 6b$ and $T-7g-7\frac g8 = T-7(g+\frac g8) = 7(-b)$
Now these two equations are Newton's second law applied in the accelerating frame of the lift with two fictitious/pseudo force equal to $6\frac g8$ and $7\frac g8$ added to the masses.
Another way of looking at this is to say that the (local = relative to the lift) acceleration of free fall is $g+\frac g8$ downwards.
If the lift was accelerating downwards at $\frac g8$ then all that would happen is that there would be a change of sign for the fictitious/pseudo force.
$T-6g+6\frac g8 = T-6(g-\frac g8) = 6b$ and $T-7g+7\frac g8 = T-7(g-\frac g8) = 7(-b)$
This is perhaps what you might expect in that if the lift is accelerating downwards at $g$ then you find that both the tension $T$ is zero and the acceleration of the masses relative to the lift is also zero.