I've encountered the following definition of spin structure:
"Let $p:P\rightarrow B$ a principal $SO(p,q)$-bundle, a spin structure" (over the associated vector bundle with metric and orientation) is a $Spin(p,q)$-bundle $\widetilde{p}:\widetilde{P}\rightarrow B$ such that
(i)There's a bundle map $\widetilde{r}:\widetilde{P}\rightarrow P$ that's a double cover;
(ii)The following diagram $$ \require{AMScd} \begin{CD} Spin(p,q)&\times & \widetilde{P}&\stackrel{right\hspace{3mm}tr.}{\longrightarrow} &\widetilde{P}\\ \downarrow{r} & &\downarrow{\widetilde{r}} && \downarrow{\widetilde{r}}\\ SO(p,q) & \times & P & \stackrel{right\hspace{3mm}tr.}{\longrightarrow} & P\\ \end{CD}$$ commutes". Here r is the natural double cover of $SO(p,q)$.
My question is about that bracket sentence :"over the associated bundle with metric and orientation". What does that mean explicitly? It seems that we're implicitly considering another vector bundle... I say that cause a spin structure over a pseudorienmanian manifold $M$ is said to be a Spin structure over the tangent bundle $TM$. What does that mean? How is the above definition linked with the tangent bundle of a manifold (that's a vector bundle)? I mean, I can't see where does $TM$ enter in the game, I hope to have been clear enough.
EDIT: Can maybe cause $TM$ has an associated frame bundle $E_{GL}$ with fiber $GL_{n}(\mathbb{R})$ which can have a reduction to $SO(p,q)$?