In short, How can we define Spin(p, q) without referencing Clifford algebras? The answer should be something like "Spin$(p, q)$ is the unique double cover of SO$^+(p, q)$ such that ...". Wikipedia seems to think we can omit the "such that ...", claiming:
Up to group isomorphism, SO$(V, Q)$ has a unique connected double cover, the spin group Spin$(V, Q)$.
[Here $Q$ is a nondegenerate quadratic form over a real or complex vector space $V$, so we can equivalently replace SO$(V, Q)$ with SO$^+(p,q)$. Wikipedia also omits the $^+$ but says that they are referring to the identity component.]
But the above claim is false. One alternative approach is to define the spin groups by explicitly constructing them with Clifford algebras, but I'd like to know of a purely topological definition, as described above.
I originally asked this question as a footnote to this one, then decided to move it here. Along with the definition requested above, I'd also appreciate some discussion of why the definition is "morally correct" or useful.
Almost nobody does this. There is a discussion in the Wikipedia article on spinor groups but it has too many mistakes and too few references. The only solid math reference I know is in Chapter 5 (freely available from author's webpage here) in
Varadarajan, V. S., Supersymmetry for mathematicians: an introduction., Courant Lecture Notes in Mathematics 11. Providence, RI: American Mathematical Society (AMS); New York, NY: Courant Institute of Mathematical Sciences (ISBN 0-8218-3574-2/pbk). vi, 300 p. (2004). ZBL1142.58009.
It is useful if you read this answer in conjunction with my answer here.
To describe $Spin(p,q)$ as a 2-fold cover of $SO^+(p,q)$ ($p>1, q>1$) one has to look at the maximal compact subgroups since they carry all the homotopy information. The group $SO^+(p,q)$ has maximal compact subgroup $SO(p)\times SO(q)$ and $$ \pi_1(SO^+(p,q))\cong \pi_1(SO(p)\times SO(q))\cong H_1\times H_2, $$ where $H_1, H_2$ are cyclic groups (either infinite cyclic or ${\mathbb Z}_2$). In particular, if $h_1, h_2$ are generators of $H_1, H_2$, then there is a homomorphism $$ \phi: H_1\times H_2\to {\mathbb Z}_2, $$ sending both $h_1, h_2$ to the generator of ${\mathbb Z}_2$. It is clear that these homomorphisms are independent of the choices of generators. Let $H< H_1\times H_2$ denote the kernel of $\phi$, it is an index 2 subgroup in $\pi_1(SO(p)\times SO(q)\cong \pi_1(SO^+(p,q))$. Then $Spin(p,q)$ is the 2-fold cover of $SO^+(p,q)$ associated with the subgroup $H$ of the fundamental group. This is your topological description of the spinor group.
In Varadarajan's terminology, here is what's going on. The (unique up to conjugation) maximal compact subgroup of $Spin(p,q)$ is $(Spin(p)\times Spin(q))/G$, where $G\cong {\mathbb Z}_2$ is generated by an element $(a,b)\in Spin(p)\times Spin(q)$, where $a, b$ are the nontrivial central elements of $Spin(p), Spin(q)$ respectively, such that $$ Spin(p)/\langle a\rangle = SO(p), Spin(q)/\langle b\rangle = SO(q). $$ Hence, $(a,b)$ corresponds to the "diagonal" element $(h_1, h_2)$ of $H_1\times H_2$.