You don't get vertical and horizontal kinetic energy, you just get kinetic energy. The same formula applies to landing on a carrier deck or a concrete runway, only the braking technique differs.
Do it all in the frame of the deck/runway, since that's where you need to end up with zero kinetic energy, unless you want a trip to the hospital.
The energy to lose in the wheel or cable braking is
0.5 x mass x (airspeed - headwind)-squared.
This assumes the headwind is in the same direction as the airspeed vector.
If not, you need to replace (airspeed-headwind) with the modulus of the vector sum of the airspeed, and the windspeed measured in the frame of the deck/runway.
If headwind=airspeed, you just latch the Harrier down without any wires, If headwind=-airspeed, you have four times the energy to lose.
Note that is not the same as Roulishollandais's
0.5 x mass x [(airspeed-squared) - (headwind-squared)].
That would give you the same braking energy advantage for a tailwind as a headwind.