Does the LHC Beam Dump experience deceleration forces, or only heat? When the proton beam is dumped into the beam dump, will the dump actually experience the deceleration force of a 200 mph freight train that is being stopped to a standstill in about a microsecond?
If yes, that must be millions of tons of force.
Or is the kinetic energy of the beam somehow converted into heat without generating deceleration forces?
 A: Force is the rate of momentum transfer, not the rate of energy transfer.
For slow objects like a 200mph train, momentum is $p\approx mv$ and kinetic energy is $T\approx\frac12mv^2,$ and so $\frac pT\approx\frac2v.$ In the case of the train, $\frac pT\approx0.01\,\mathrm{mph}^{-1}.$
For fast objects like the LHC protons, momentum is $p=\gamma mv$ and the kinetic energy is $T=(\gamma-1)mc^2\approx\gamma mc^2$ (where $\gamma=1/\sqrt{1-\frac{v2}{c^2}}$ shows up due to special relativity), and so $\frac pT\approx\frac v{c^2}.$ Approximating further $v\approx c,$ $\frac pT\approx\frac1c.$
Comparing these formulae, the LHC proton beam has only $\frac{1/c}{0.01\,\mathrm{mph}^{-1}}\approx0.00000015$ times the momentum as the "freight train of equivalent energy". The beam might have the energy of a train going 200mph, but it has the momentum of one going $0.00000015$ times as fast, or less than 2 inches per hour. Thus, dumping the beam requires handling a large amount of energy but doesn't produce nearly as much recoil as you are expecting. This is done (per the article) by heating a large mass that doesn't move significantly.
Note that SR is not required to get this qualitative result; without SR the $1/c$ would simply be $2/c$ and the momentum estimate would be (incorrectly) doubled. The fundamental intuition that is valid both relativistically and nonrelativistically is that as something speeds up, its momentum grows slower than its kinetic energy. For something going as ludicrously fast as an LHC proton, you can have a very large amount of energy without much momentum.
