Is there a relation between spin and the spin group?

In quantum mechanics, the relevant representations of symmetry groups on the space of states are not our usual linear representation, but projective representations on the Hilbert space. The projective representations of a semi-simple Lie group - such as the rotation group $\mathrm{SO}(n)$ - are in bijection to linear representations of its universal cover. For a detailed discussion and a derivation of these facts, see this Q&A of mine. The "intuition" for the appearance fo projective representation is that states really are not vectors but rays in Hilbert space and hence "phases don't matter".

Now, the rotation group in $n > 2$ dimensions has fundamental group $\mathbb{Z}/2\mathbb{Z}$, meaning its universal cover is just a double cover. Therefore, in $n>2$ dimensions, the spin group is by definition its double cover and hence the group we need to linearly represent on the Hilbert space to have a projective representation of the rotation group. Our beloved half-integer "spin" $s$ is now nothing but the number uniquely labeling an irreducible linear representation of $\mathrm{Spin}(3)$ by $L_x^2 + L_y^2 + L_z^2 = s(s+1)$.

The spin group is related to spin-half objects, called spinors. If you rotate a spinor by 360 degrees, you get back the negative of the spinor you started with. Now it would be nice if you could represent the action of this rotation by saying that an element of $SO(n)$ is acting on the spinor. However, this cannot be done because a rotation by 360 degrees is the same as the identity element of $SO(n)$, and so the action of this rotation must be to leave the spinor invariant, contrary to what we know happens. Thus there is no way to accurately represent the action of $SO(n)$ rotations on a spinor.

However, if you had a bigger group, where a rotation of 360 degrees didn't take you back to the identity element, then you might be able to make a one to one correspondence between elements of this bigger group and the linear transformation it causes on spinors.

The spin group is this bigger group. Since the spin group is a double cover of $SO(n)$, a rotation of 360 degrees only takes you half way around the spin group, and so the group element corresponding to a 360 degree rotation is not constrained to act as the identity on the spinor, but instead can multiply the spinor by $-1$, as it should.

So to summarize, you can make a group out of all the finite rotations that can be applied to a spinor. This group is not $SO(n)$, since the identity rotation and a 360 degree rotation act differently on the spinor but are the same in $SO(n)$. So the group of finite rotations that can be applied to a spinor must be bigger than $SO(n)$. In fact, this group turns out to be the spin group, which is a double cover of $SO(n)$