The structure group of $TX \oplus T^*X$ is $SO(n,n)$. Since $w_2(TX) = w_2(T^*X)$, this bundle always admits a spin structure and we lift to $Spin(n,n)$. The sections of the associated spinor bundle are not spinors in the usual way but bispinors like you guessed (they have two spinor indices, one from $TX$ and one from $T^*X$).

A B-field (in the sense of http://arxiv.org/pdf/math/0401221v1 ) determines an extension

$T^*X \to E \to TX$.

By the splitting principle, this bundle also admits a sort of bispinor.

Now it is a bit confusing to me since in physical applications (eg. the wonderful thesis of Yi Li: http://thesis.library.caltech.edu/2098/) only ordinary spinors are used, though I think that the (complex) bispinors are $\psi_+ \otimes \psi_-$ and $\bar\psi_+ \otimes \bar\psi_-$. We have $Spin(n,n) = Spin(n) \times_{\ZZ/2} Spin(n)$ and accordingly the spinor representation splits as a tensor product. A spin structure on $TX$ I think gives a basis of the sections of the bispinor bundle of simple tensors, so we get well-defined $\psi_+$ and $\psi_-$.