It is relatively simple. If you integrate the DOS (from -infinity to the the Fermi energy) in both, the majority and minority spin sectors , their difference gives the global magnetic moment per unit cell.
If you have the DOS resolved in species (local DOS) then you can obtain the magnetic moment per type of atoms (or orbitals).
However, you will not get any information regarding the couplings between the localized moments from the DOS.
As per the data you have shared, first look at the Fermi level, which is 10.664 eV.
When you will plot this data, you will be able to see that both spin up as well as down are crossing the Fermi level, so the system is metallic.
and you can also see that there is a finite difference between spin up and down states, which means that the system has a finite magnetic moment, and is Ferromagnetic in nature. and the net magnetic moment will be the difference of spin-up and spin-down charges. This information you can also get this from the output of your QE calculations.
if both spin-up and down states would have equal contributions to DOS, then the system would have been anti-ferromagnetic.
For more details on couplings of different atoms and their orbitals, you should plot the projected density of states.
Probably this is because, in ordinary DFT all calculations are done assuming absolute zero temperature. Therefore, all states above the Fermi level are completely empty. So, integration above Fermi level is not necessary.
However, for TD-DFT calculations or for suitable perturbations, this statement will not be valid anymore.