Thanks for the reply Tobias. Wouldn't $F_{x}$ be the space of oriented and pseudo-orthonormal frames of $T_{x}M$, just to be a little more precise? (Since $T_{x}M$ has Lorentzian metric signature). And I understand the $Spin(1,1)$-bundle now, thank you.

Sorry for resurrecting this discussion. Just to make sure I understand the upshot of all this: Do the spinors $\psi_{+}$ and $\psi_{-}$ correspond to the sections of the spinor bundles $S=K^{1/2}$ and $S=\bar{K}^{1/2}$ respectively? Whilst $\psi^{i}_{+} \dfrac{\partial}{\partial \phi^{i}}$ and $\psi^{i}_{-} \dfrac{\partial}{\partial \phi^{i}}$ are sections of $K^{1/2} \times \phi*(TX)$ and $\bar{K}^{1/2} \times \phi*(TX)$ respectiveley?

Just to add to this discussion: I think the importance of the transformation being unitary might be the fact that unitary transformations don't alter the integration measure. So before the Fourier transformation we had $\mathcal{D} \phi(x)$ and after the transformation we still have $\mathcal{D} \phi(x)$, it just so happens that this can be expressed in terms of real and imaginary $\phi(k)$. Non-unitary transformations may have altered the integration measure so that $\mathcal{D} \phi(x)$ can't be used. It would be helpful if somebody could either confirm or reject this reasoning however.

Ok, I think I get it now! In the first diagram I drew assume both the incoming and outgoing p's are from left to right. It would be perfectly consistent to change the arrow of the outgoing such that it points from right to left so long as it's also relabelled -p. Thus both arrows are incoming but momentum is still conserved. Is this right?

I see what you're saying, but would you not have to stick to the same choice of direction for both the incoming and outgoing momenta (such that both would be p or both would be -p). It seems odd (cheating?) to change the direction arbitrarily from one part of the diagram to another.

Ah I see, so is it correct that the first diagram I drew isn't complete, it's just the internal part of another (arbitrary) larger diagram? Would this be true also for the diagram that gives equation (9.80) on page 305 of Peskin and Schroeder? Also I don't completely understand how both momenta can point inwards without violating momentum conservation. I would have thought the momentum 'going in' has to equal the momentum 'going out'. That is to say, I would have though both gauge fields would need the same momentum (p for both).