Lifting a diffeomorphism into a spinor bundle automorphism

As spinors transform with a minus sign under a full rotation, there is no (non-trivial) lift of the action of the group of diffeomorphisms to the spinor bundle (i.e. the spinor bundle is not a natural bundle). This is closely related to the fact that a spinor transforms according to a projective representation of the Lorentz group.

Nonetheless, you can lift vector fields to the spinor bundle (using a connection). Hence, the spinor bundle is an infinitesimally natural bundle and this suffices to construct the stress-energy-momentum tensor. For more details about infinitesimally natural bundles, see Bundles with a lift of infinitesimal diffeomorphisms or Infinitesimally natural principal bundles.

There is an intrinsic ambiguity in lifting diffeomorphisms to spinors, since as you say spinor bundles are not natural bundles. But the ambiguity is not large and has more to do with the global properties of spin structures.

First, recall that the oriented orthogonal frame bundle $\mathcal{P}_g M$ of a pseudo-Riemannian manifold $(M,g)$ is an $\mathrm{SO}(p,q)$-principal bundle, where $(p,q)$ represents the signature of the metric $g$. Then recall that a spin structure is a $\mathrm{Spin}(p,q)$-principal bundle $\mathcal{S}M$ together with a principal bundle morphism $\mathcal{S}M \to \mathcal{P}_gM$, which fiberwise restricts to the usual projection homomorphism $\mathrm{Spin}(p,q) \to \mathrm{SO}(p,q)$. Now, given a linear representation of of $\mathrm{Spin}(p,q)$ on a vector space $\Sigma$, in particular a spinor representation, the associated vector bundle $\Sigma_{\mathcal{S}} M = (\mathcal{S} M \times \Sigma)/\mathrm{Spin}(p,q)$ is the corresponding spinor bundle over $M$.

Now, a diffeomorphism $\phi\colon M \to M'$ (at least one that preserves metrics and orientations of oriented pseudo-Riemannian manifolds $(M,g)$ and $(M',g')$, which is all that we will consider for now) induces a morphism of oriented frame bundles $\phi^*\colon \mathcal{P}_g M \to \mathcal{P}_{g'}M$. This is a principal bundle morphism, meaning that it is $\mathrm{SO}(p,q)$-equivariant. Suppose that there exists also a $\mathrm{Spin}(p,q)$-principal bundle morphism $\tilde{\phi}^*\colon \mathcal{S}M \to \mathcal{S}'M'$ of spin structures that covers $\phi^*$. The existence of $\tilde{\phi}^*$ is not automatic, possibly it doesn't exist at all. But if it does exist, then it is not unique, since the $\mathbb{Z}_2$ center of $\mathrm{Spin}(p,q)$ acts non-trivially on $\tilde{\phi}^*$. But that's the only freedom you have; if $\tilde{\phi}^*$ exists, then there exist exactly two possibilities. Basically, it is sufficient to fix the lift of $\phi^*$ at a single point $x\in M$ (there are only two possibilities) and the rest is determined by continuity.

Finally, all that remains to observe is that a morphism of principal bundles induces a morphism of associated bundles. Hence, a spin-lift $\tilde{\phi}^*$ of our diffeomorphism $\phi$ induces a vector bundle morphism $\tilde{\phi}^*_\Sigma \colon \Sigma_{\mathcal{S}} M \to \Sigma_{\mathcal{S}'} M'$. This morphism can be constructed as follows. First, consider the map $$ (\tilde{\phi}^*, \mathrm{id}) \colon \mathcal{S}M \times \Sigma \to \mathcal{S}'M' \times \Sigma . $$ Here, the OP's original intuition that spinors should be treated as "scalar fields" whose pointwise values are unchanged by push-forward along diffeomorphisms is implemented by the fact that the above map acts as the identity on the $\Sigma$ factor. Next, note that $(\tilde{\phi}^*, \mathrm{id})$ is $\mathrm{Spin}(p,q)$-equivariant, and hence induces a map of the associated bundles, which is precisely our desired $\tilde{\phi}^*_\Sigma$ push-forward of spinor fields.

It remains only to recall the OP's final piece of intuition, that spinor fields transform as a "spinor" under changes of orthogonal frames and see how it is implemented in the above construction. Locally, a spin structure is often specified by a local choice of a specific frame field. Say that we have chosen specific frame fields $e^i$ on $(M,g)$ and $e'^i$ on $(M',g')$. Obviously, upon pushing forward $\phi^* e^i = \Phi^i_j e'^j$, where $\Phi$ is an $\mathrm{SO}(p,q)$ valued function, which need not be the identity. So, if $\sigma^\alpha(x)$ is a spinor field, that is, a section of $\Sigma_\mathcal{S} M$, then the push-forward $\sigma' = \tilde{\phi}^*_\Sigma \sigma$ spinor field is defined by the formula $$ \sigma'^\alpha(\phi(x)) = \Phi^\alpha_\beta \sigma^\beta(x) , $$ where $\Phi^\alpha_\beta$ is the matrix representing the pointwise action of $\Phi$ on $\Sigma$. The rigid frame rotation (or "Lorentz transformation") $\Phi$ takes into account the difference between the orthogonal frame $e'^i$ on $(M',g')$ and the push-forward $\phi^* e^i$ from $(M,g)$.

I have one more small remark to make about the meaning of the components $\sigma^\alpha(x)$ on a manifold. On flat space, they would be components with respect to some basis in the vector space $\Sigma$. But, on a manifold, these components obviously need to be interpreted with respect to some (spin-)frame in the spinor bundle $\Sigma_{\mathcal{S}} M$, but a global frame need not always exist, and even locally there is huge freedom in choosing in choosing a spin-frame as a function on $M$. What is important for the above push-forward of spinor fields to make sense, the spin-frames on $\Sigma_{\mathcal{S}} M$ and $\Sigma_{\mathcal{S}'} M'$ need to be chosen such that the $\gamma$-matrices which implement the Clifford multiplication and the action of $TM$ on $\Sigma_{\mathcal{S}} M$ must be fixed. That is, there must be some constant matrix $(\Gamma^i)^\alpha_\beta$ such that $(\gamma^i)^\alpha_\beta(x) = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M$, while at the same time also $(\gamma'^i)^\alpha_\beta(x') = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M'$. So the above push-forward rule only works when considering spin-frames of this kind.