am I right in thinking that the coefficients [itex]b_{ij}[/itex] in the quote above are not [itex]b(e_i,e_j)[/itex], but rather [itex]b_{ij} = 2b(e_i,e_j)[/itex]?

By [itex](e_i)[/itex], I mean the basis for [itex]V[/itex] with dual basis [itex](\varepsilon^i)[/itex].

Ah, no, looking again at this, I think they must mean the usual coefficient [itex]b_{ij}=b(e_i,e_j)[/itex] after all.

A factor of 2 would be needed to convert components with respect to a tensor product basis [itex](\varepsilon^i\otimes\varepsilon^j)[/itex] for [itex]T^0_2[/itex] (all permutations of the indices) to the corresponding wedge product basis [itex](\varepsilon^i\wedge\varepsilon^j)[/itex] (only increasing permutations) for [itex]\wedge^2 V^*[/itex]. But in this case, I think, they're not restricting the indices to only increasing permutations.

So, for example, if [itex]\text{dim} V = 3[/itex] and [itex]b=b_{12} \varepsilon^1\otimes\varepsilon^2 + b_{21} \varepsilon^2\otimes\varepsilon^1[/itex], where [itex]b_{12}=-b_{21}[/itex], we have