If a charge is moved from lower power potential to higher potential, then energy should be

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What you understand has to be refined a bit. Potential is defined referring to what happens to a positive charge in a field. A positive charge is moved by the electric field from the higher to the lower potential. Therefore, a negative charge is moved in the opposite direction. See the diagram:

Now to your formula:

Indeed, for a positive charge in an electric field

$ (1) \ \Delta V = V_{final} - V_{initial} < 0.$

Pay attention that the electric potential energy is equal to

$ (2) E_{potential} = qV .$

Therefore,

$ (3) \ \Delta E_{potential} = q \Delta V < 0.$

The electric force accelerates the positively charged body increasing the body's kinetic energy, on the expense of the body's potential energy which decreases.

Now, for a negative charge the forces move it from the lower potential to the higher one. Nothing wrong happens. According to the inequality (1), $\Delta V > 0$, however, due to the definition (2), the equality (3) gives again $\Delta E_{potential} < 0.$ because $q$ is negative. So, again the electric force accelerates the body increasing its kinetic energy on the expanse of the potential energy of the body, which decreases.

Let me assume for simplicity that we have some big charge(s), and we bring a positive test charge. So, there is a field between the big charge(s) and the test charge. The field encapsulates energy. (For simplicity I will also assume that the big charges are placed on massive bodies.)

Now, for out test-charge there are two situations of passing from one value of the potential energy to another value, and in both is done work.

A) When the charge passes from a higher potential to a lower potential, work is done, but it is done by the force of the field. The field looses energy but this energy becomes kinetic energy $KE$ of the test charge. (To be rigorous, the field forces move all the charges not only your test charge, but for avoiding this complication I assumed that the big charges are placed on massive bodies, s.t. the velocities they get may be neglected.)

So, in the case when field energy (potential energy) transforms into kinetic energy of the test charge, the work is done by the field forces.

B) When we raise a test charge from a lower potential energy to a higher potential energy, the field energy increases. If the test charge had kinetic energy $KE$, it decreases. But if it was at rest, and we want to displace it to a higher energy, it's we that have to do the work. We invest an amount of work $W = q\Delta U$ and this energy is absorbed by the field.