Sure, they aim to extract energy directly from the field, but the three breakeven points are still important. A significant part of the energy will be lost as x rays and neutrons, since their D-H3 fuel cycle is not aneutronic; they will also have significant D-D reactions that are required to breed Tritium which they capture and then let ti decay to Helium-3.
Overall, when you look at the total complexity and energy balance of the full reactor + fueling cycle, maintaining vacuum, keeping superconducting magnets at cryogenic temperatures, tritium extraction etc. then generating an order of magnitude more energy than inserted still seems necessary to achieve engineering breakeven.
In the cycle under question (two DD reactions per D3He reaction), 91% of the fusion energy goes into charged nuclei, not into neutrons. In steady state where T is being allowed to decay into 3He and there's just one DD reaction, the fraction of energy in charged nuclei is even higher.
X-ray emission is strongly dependent on electron temperature. One of the important aspects of Helion's scheme is the electron temperature is much lower than the ion temperature. Not only does this greatly reduce x-ray emission, it reduces plasma pressure at a given ion temperature vs. a plasma where the ions and electrons are in thermal equilibrium, thereby increasing the ion density and fusion rate. The pulses in Helion's scheme end (and the plasma energy is recovered) before the electrons can heat up.