Are spinors, at least mathematically, representations of the universal cover of a lie group, that do not descend to the group?

Yes, yes, and yes.

In more detail: issues about smooth vectors are much subordinate in any case, and prove trivial, so (smooth-vector?) repns of the "lower" group "lift" to the covering group. Not at all a problem.

Yes, at least as evidenced by many examples, about half the finite-dimensional repns of spin groups do not descend to the corresponding orthogonal groups, for example. Some other classical groups happen to be simply-connected, in contrast.

And, yes, to the last question, as a matter of usage or convention or tradition.

The underlying point-of-interest is that it is not obvious that the universal covering of all these not-simply-connected Lie groups is just a two-fold-cover. Weyl found this.

In quantum mechanics, complex state vectors are always defined up to a "phase". What matters is the direction of a vector, any given $\psi$ is physically equivalent to $z\psi$, where $z$ is a complex number (of norm $1$ for probability normalization). That, is, physics lies in a complex projective space.

A linear representation induces always a projective representation, of course. The converse is not always true. But it is true that a projective representation induces a linear representation, in general, of a larger group, called "central extension". (If you know what a central charge is, it is exactly that, in mathematical language.) For Lie groups, extensions by a discrete group are covering groups. If the group in question is simply connected, then, this problem is trivial.

$SO(n)$ is not simply connected. So there are projective (or physical) representation of $SO(n)$ that cannot be written as linear representations, but still have physical significance.

In a way, spinors are representations of the orthogonal group. Only, not linear.