Is the quickest way to reverse direction by turning horizontally or vertically?
The formulae appear correct, as far as they go. But there are some problems with the underlying assumption of the question; specifically, that you can achieve either condition in a steady-state fashion.
Firstly, and most importantly, you can see looking at both formulae, the most significant thing you can do to improve your turn rate (angular velocity) is to slow down. This also has a significant effect on the turn radius as well. Since the scenarios where one cares about maximum performance turn rate often also involve a need to avoid an obstacle (cumulonimbus, cumulogranite, etc.), minimizing the turn radius can be just as important as maximizing turn rate, and both can be accomplished by drastically reducing speed.
Conveniently, one of the things that makes it difficult to compare the two scenarios is the fact that the moment you begin the turn, the increased load factor will cause an increase in drag, which will help reduce the speed. The speed reduction (which may be aided by judicious reduction of power) can dramatically improve turn performance, but this reduction will depend on exactly how that turn is made. It may even involve an increase in speed, if attempting the looping turn by performing a split-S.
Importantly, you are trying to compare apples-to-apples by keeping the load factor identical in both scenarios, but I don't find that a plausible assumption. It's not too hard to maintain a constant load factor during a level turn, but a looping turn is likely going to involve either increasing-then-decreasing, or decreasing-then-increasing load factor, making the assumption of load factor being the same in both scenarios unreasonable. And in either case, the speed will be changing by a large amount, eliminating the possibility of using those formulae in an apples-to-apples manner.
Finally, there is the issue that many planes cannot be safely looped, and even among those that can, you may not be able to initiate a loop (Immelmann or split-S) from cruise speed. You have set the assumption that you are cruising at Mach 2, which seems to be an attempt to address this concern. But it raises new ones: even an airplane capable of traveling at Mach 2, may only be able to fly that speed when at altitude, and at an angle-of-attack close to its critical angle of attack, making an Immelmann loop impractical, if not impossible. Similarly, it does not seem likely that one could safely accomplish a split-S that is initiated at Mach 2. The altitude loss incurred could easily exceed the altitude the airplane is starting the turn at, and even if not, the increased airspeed from Mach 2, as well as the G load required at the bottom of the loop, could easily exceed the airplane's structural limits (or the pilot's!).
The conclusion of your question is particularly confusing:

my LIFT FORCE (no matter where it acts on my aircraft) is the limiting factor, nothing else. Is that correct?

and

lift should not be responsible for a quick turn but the lever arm?

Indeed, during a level turn, with a large enough bank angle, one can improve turn performance at the expense of added drag and reduced speed by pulling on the elevator, i.e. adding nose-up elevator input.
But I'm not sure that's what you're asking. It sounds like maybe you're wondering about using lift to modify the attitude of the airplane. If so, that's not how it works. While it's true (for a conventional airplane configuration) that whether in level flight or not, the pitch of the airplane, controlled by the elevator input, changes around the center of gravity, it's important to keep in mind that the center of gravity in unaccelerated flight is directly below the center of lift, since the elevator has effectively added weight, shifting the CG rearward.
I.e. there's no &quot;lever arm&quot; related to any distance between center of gravity and center of lift.
But more to the point, changing the attitude of the airplane isn't the same as changing the flight direction of the airplane. Taking this to its extreme, you can imagine an airplane that could somehow yaw 180 degrees instantaneously. Such an airplane would not effect a 180 degree turn instantaneously; instead, it would just wind up flying the same direction it had already been going, just backwards.
The bottom line: the formulae you're using do not appear to be amenable to an apples-to-apples comparison, so the academic line of reason doesn't seem resolvable. And in the practical sense, if you're going Mach 2 and actually need a maximum performance turn, one of the first things you need to do is slow down. It is most likely safest and most practical to accomplish this in a level turn, or possibly some combination of bank and climb (e.g. like a chandelle). Either way, it's likely the main limiting factor will be the maximum load factor that the airplane and pilot are capable of sustaining.
I don't think it's possible, given only and based solely on the formulae and conditions stated in your post, to determine whether it's always true that a level turn or either type of looping turn is going to produce the maximum rate of turn, first and foremost because you're never going to keep the velocity term in the formulae constant, but for other reasons as well.