Because we use magnetic fields to bend the path of particles in accelerators and E&M is Lorentz invariant by construction, we just apply the bending radius in a magnetic field equation in the lab frame and never bother to compute the force. The radius of curvature is
$$ R = \frac{p}{qB} .$$
Note that for a ring like the LHC, the bending is not actually uniform, but only in the bending magnets, but not in the quadrupoles or cavities (if any), so it is locally tighter than you would get by naively apply the above equation with a 27 mile radius.
If you insist on finding the force you'll note that over the course of one cycle the momentum changes by $2\pi |p|$, and it takes $(27\text{ km})/c$ seconds to get there so the mean force is
$$ F = \frac{p c}{r} = \frac{2 \pi p c}{27 \,\mathrm{Km}}. $$
Again, in any given magnet it will be larger by a factor of less than 10 because the magnets don't cover the whole beam line.
Aside: If you spend much time doing particle physics you'll come to love ultra-relativistic mechanics: it's even easier than non-relativistic mechanics.
Isn't the universe full of Higgs bosons, making up the Higgs field?
No. In particle physics, it is understood that the underlying (more fundamental) object is the field, not the particles. Particles are excitations of the fields that can be measured, and always carry certain properties like charge, mass, spin etc. The field that you are most familiar with is the electromagnetic field, its excitations being the photons. In another field the excitations are electrons, in another still there are gluons etc.
And there is a Higgs field, whose excitations are the Higgs bosons. The Higgs field, in contrast to the electromagnetic field has a non-zero value even if there are no Higgs bosons there.
To have an analogy in mind, think of a room full of air. When I speak, there are sound waves moving around the air. The air is the Higgs field, the sound waves are the Higgs bosons.
Why can't we detect the ones that are there already, like we can other bosons such as photons?
Higgs bosons are very massive, as particles go, so they require a lot of energy to be created in collisions. Additionally they have a number of decays pathways, so when they are created, they decay rapidly. So, even if Higgs bosons are created all the time in the atmosphere, or in supernovae or other events, they are rare and hard to detect. That is why we set up an experiment that can reproduce millions of collisions a second so to accumulate enough data.
When we've made our Higgs out of pure energy, why does it instantly decay into other particles?
This is kind of misguided. There is no clear meaning of "pure" energy. Energy is a quantity that is assigned to various phenomena, yet is common and interchangeable between them all. We speak of kinetic energy, potential energy, mass-energy, etc. but none of these forms is "purer" in any specific sense. In the particle collisions, the kinetic and rest mass energy of the protons is concentrated in a small part of spacetime, and can be redistributed in the kinetic, potential and mass energy of other particles.
Once a particle is formed, it does not really matter what way it has been formed. Just like a radioactive nucleus has the same probability of decaying in the next $10$ minutes irrespective of how long it has survived until now, a Higgs boson will decay with a certain probability into the particles it can decay to.
Does it actually directly decay into other particles, or is it rather the case that it just turns back into pure energy, and then that energy produces other, less massive particles?
Here we end up a little in metaphysics.You will have different answers depending on the interpretation of QM you choose. All we observe is the protons that go in the collision, and the shower of particles that comes out after the collision, together with their energy. That's all. Quantum theory will give you the statistics of these observations, but not what happens between the two observations; that is (for now) metaphysics, because it is unobservable.
Strictly speaking, no Higgs boson has been observed, in the sense that no Higgs boson has collided with the detectors. We have calculated how the existence of the Higgs field will affect the measurements, we found that it would affect them in a particular way, we did the experiments and indeed found that signature. The experiments and theory match so well that it is inescapable that there is a Higgs field, even though we have not "seen" (with our eyes) any Higgs bosons.
To speak about the exact way in which one particle comes into existence and decays is a bit beyond present physics (also worth exploring in other questions).
Best Answer
Since the energy of a particle is frame dependent, it is not particularly meaningful to discuss the energy of a single particle, by itself. The reason for this is that one can find a reference frame where that particle has any arbitrary kinetic energy, and since the laws of physics are the same in all inertial reference frames, we cannot have a proton disintegrate in one frame and continue existing in another. This is why the protons must collide with something else (another proton moving the opposite direction, in the case of the LHC) in order to produce an interesting reaction. As such, the more interesting quantity to look at is the energy of the collision, which is the same in all reference frames, given by the sum of the energies of the protons in the centre of mass frame.