i do not understand the part of clamping the center of the coil or
cutting it off
The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer.
If you have taken a first semester in electrostatics, you may be aware of the method of images. In that case, there isn't a symmetry but we create a symmetric problem with a known solution by extending the domain with a mirror image.
This is essentially the reverse of that approach. We have a symmetric problem which we solve by appeal to the known solution of the unsymmetric problem by cutting the domain in half.
Since the problem is symmetric, we know that if one mass is displaced by $\Delta x$, the other mass is displaced by an equal and opposite amount.
Thus, the length of the spring changes by $2\Delta x$ so the magnitude of the force on either mass changes by $2k\Delta d$
The fact is, and this shouldn't be hard to convince yourself of, that if we only look at 'one side' of this problem, it appears that either mass is attached to a spring that is fixed at the center and has spring constant $2k$. Thus, the frequency of oscillation of this 'one sided' problem is
$$\omega = \sqrt{2k/m} = \sqrt{2}\sqrt{k/m}$$
Appealing to symmetry is a powerful method to 'intuit' an answer from known solutions.
As regards to the effect of gravity, assuming that the spring is linear, the frequency of oscillation is unaffected by a constant offset force.
That is to say, the frequency of oscillation, $\omega = \sqrt{2k/m}$ is the same for the following two differential equations:
$$m\ddot x + 2kx = 0$$
$$m\ddot x + 2kx = mg$$
The effect of the constant gravitational force on the right hand side of the 2nd equation is essentially to add a constant offset to the solution for the position function without changing the frequency.
The charge on one block not only repels the charge on the other block, but it also repels the $dq$ charge that you are bringing to charge the system.
As you are slowly charging the system and slowly building up the charges, the force $F_e$ as you integrate is not a constant since the charge is increasing slowly. You may also need to account for the self-energy to build the charged system from scratch (a charge of 0 to 2Q).
Best Answer
Irrespective of the end to which a block I attached. The reading of the spring balance will always be equal to the tension in the spring balance.
When blocks attached are of equal masses the tension would be equal to the weight of one block.
In case of unequal masses attached on either ends, the masses will undergo acceleration and you can calculate the tension using Newton's Third Law.