Let $(M,g)$ be a semi-Riemannian manifold with metric of signature $(p,q)$. I believe the signature of the metric is not relevant for this discussion so I leave it arbitrary (corrections to this are highly appreciated!).
Now let $\xi = (F(M),\pi,M,{\rm SO}(p,q))$ be the principal bundle of orthonormal frames over $M$. Suppose that $M$ admits a spin structure and that $\zeta = (P,\pi_P,M,{\rm Spin}(p,q))$ is the associated principal bundle of spin frames and $\Phi:P\to F(M)$ is the principal map obeying $\Phi(p\cdot g)=\Phi(p)\cdot \rho(g)$ where $\rho: {\rm Spin}(p,q)\to {\rm SO}(p,q)$ is the covering map and where the $\cdot$ is the right $G$-action on both $P$ and $F(M)$.
Now it is clear that if the spin frame bundle $\zeta$ admits a global section $\sigma:M\to P$ then so does $\xi$. In fact $\Phi\circ \sigma : M\to F(M)$, it is a section because $\Phi$ is a fiber bundle morphism. Moreover if $\sigma$ is a smooth section, assuming $\Phi$ to be smooth, so will be $\Phi\circ\sigma$.
Now my question is about the converse. If the orthonormal frame bundle $\xi$ admits a global section $E:M\to F(M)$ can we prove that the bundle of spin frames $\zeta$ will also have a global section? Obviously we would like to define $\sigma:M\to P$ by $$\Phi(\sigma(x))=E(x),\tag{1}$$
but the issue is that $\Phi$ is not injective. In fact it is $2$-$1$. In that setting for each $x\in M$ there are two possible solutions for $\sigma(x)$ via (1). We can clearly pick one of such solutions at each point and indeed construct a map $\sigma:M\to P$ with $\pi_P\circ \sigma= \operatorname{id}_M$ but can we do so in order that the resulting $\sigma$ is continuous or smooth?
I have the intuition that the answer is yes and that it is very basic and I'm missing it, but I really don't know how to argue this rigorously.