Total spin number 0 together with $B_2$ state - confusion
We can see what a $^1B_2$ will look like by consider the MO diagram for water, a $C_{2v}$ molecule with $10$ electrons.

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This is the configuration that we might expect for the ground state, which has $^1A_1$ symmetry. We could get a configuration with $^1B_2$ in a couple ways, namely by exciting form $1b_2$ into $4a_1$ or by exciting from either $2a_1$ or $3a_1$ into $2b_2$ without changing the spin of whatever electron we move. These new configurations would still be singlets because we haven't changed any of the spins and would have $B_2$ symmetry since clearly $B_2\otimes A_1=B_2$. 

As to your question about spin multiplicity, if we consider a system with a single electron, we account for both the spin up and spin down configuration by determining the spin multiplicity. The spin multiplicity for $S=\frac{1}{2}$ is $2S+1=2$, a doublet, which gives us exactly the number of spins states we would expect.