I don't know if that's simple enough, but in QM spin is not an ordinary vector. In general, in QM when you have two systems (say two particles) represented by [itex]|\psi_1\rangle[/itex] and [itex]|\psi_2\rangle[/itex], the total state composed by the two particles is given by their tensor product: [itex]|\psi_{tot}\rangle=|\psi_1\rangle|\psi_2\rangle[/itex].
The same thing is valid for spins. If you have two particles with, for example, spins [itex]S_1=1/2[/itex] and [itex]S_2=1/2[/itex], then the total spin [itex]S=S_1+S_2[/itex] is given by the tensor product [itex]|S\rangle=|S_1,S_{1z}\rangle|S_2,S_{2z}\rangle[/itex], where [itex]S_{iz}[/itex] is the value of the third component of the i-th spin. To do this composition you need the so called Clebsch-Gordan coefficients. Anyway, the final spin can be [itex]S=0,1[/itex] and the value of its third component is given by this tables.
This is very very shortly explained but I hope is clear enough.