Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system.
I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on the surface of sphere they most definitely exert net force on the charge $Q-q'.$
But why doesn't $\left(Q- q'\right)$ exert force on $q'$ on the surface??
It does. The charges that make up the density of charge $Q-q'$ exerts forces on each other so they exert force on the charges that make up the density of charge $q'$ in exactly the same way.
On the other hand, superposition principle tells that you can't alter the configuration of either system when superimposing the two systems which means the surface charge density of $q'$ remains the same even when there are $\left(Q- q'\right)$ other charges on the surface??
Let's be straight, when you grounded the sphere the external charge $q$ felt a force as if there were a charge $q'$ at the image charge location. And there is a net charge of $q'$ on the surface and the charges that make up the net charge of $q'$ exert forces on each other but the forces on each other add up to zero. But they still feel a force due to the charge $q$ and in fact the total force on all the charge on the surface is equal and opposite to the force the charge $q$ feels due to an image charge at the image location. And the actual force at each point in the sphere points orthogonal to the surface.
That orthogonality is the sole fact we get from the fact that the surface is an equipotential surface. And the force that is orthogonal is the total force i.e. the force due to the other charges on the surface plus the force due to the charge $q$ ... only their sum is orthogonal.
As for superposition, the force due to the uniform spread of charge $Q-q'$ over the surface is a force that points radially outwards as well.
The densities add, the fields add and the forces add. The only reason things don't move is that forces orthogonal the surface don't allow charges to move.
So, why doesn't $\left(Q- q'\right)$ exert force on $q'$??
It does.
If it exerts force on $q'$, then it would change the distribution of $q'$
No it doesn't, because the net force is orthogonal to the surface. And charges are free to move inside a conductor but they are not free to move outside a conductor. So a force orthogonal to the surface simply doesn't move the charges because the conductor itself is capable of keeping charges from leaving the conductor. It is called a work function.
which can't happen as said by superposition principle.
And now I think you misunderstand the superposition principle. It just says that if you add fields and add charges that it is also a solutions. And it does mean that if you add static solutions, you get static solution. If 100% everything were electrostatics then yes you could add like that. But consider the simple case of adding charge $q''$ uniformly spread to the outside of a conducting sphere, no grounds, no external charge. If you kept adding solutions like that over and over again eventually the work function might not be enough to keep such a huge charge density on the surface. Or maybe the field density outside the sphere gets large enough to break down the dielectric that is the air itself outside the sphere.
Just because you add charge and current and fields together doesn't mean that things don't move just because they didn't move on their own. Because the sphere itself or the air itself has to deal with all those things added together and it could have limits to what it can handle.
Superposition is just having a linear equation, add the sources and add the individual solutions and you get a solution. Anything non-linear such as a work function or a dielectric breakdown means the effects might not be static even if the individual things were static. Superposition doesn't mean more than it means.
Why did then Jackson write that the force on $Q - q'$ from $q$ is balanced by $q'$?
The forces due to $q$ and $q'$ are already balanced in the sense that the total field due to both is orthogonal to the surface so the new charge also arranges itself so that its field is also orthogonal to the surface as well which means the new charge is arranged uniformly.
