I do not understand the difference between SO(3,1) and Spin(3,1) I understand the abstract theoretical notion that Spin(3,1) is a double cover of SO(3,1), but I cannot process this when it comes to my choice of representation.
I am using the following representation of SO(3,1)
$$
R=\exp ( \frac{1}{2}B)
$$
where B is a bivector of clifford algebra over $\mathbb{R}^{3,1}$.
Do I have a representation of SO(3,1), or of Spin(3,1).
If I only have SO(3,1), what is missing from my representation to get Spin(3,1)?
 A: The bivectors are elements of the Clifford algebra which are linear combinations of those of the form $uv$ with $u$ and $v$ perpendicular (WLOG you could pick $u=e_i,v=e_j$ standard basis vectors with $i<j$). When you exponentiate them, you're plugging them into a power series, and the result is once again an element of the Clifford algebra, more specifically in the spin group which resides inside the algebra.
For example, consider $e_ie_j$. You can verify $R=\exp(\theta e_ie_j)=\cos(\theta)+\sin(\theta)e_ie_j$. If you define the spin group as being generated by products of evenly-many unit or pseudo-unit vectors (those which square to $\pm1$ in the algebra, meaning their (pseudo)-Euclidean squared norm is $\pm1$), we can see this $R$ is in the spin group by writing it as $e_i(-\cos\theta e_i+\sin\theta e_j)$.
The vector space is considered a subspace of the Clifford algebra. (The algebra is graded, starting with scalars at grade $0$, then vectors at grade $1$, then bivectors, etc.) And it is closed under the conjugation-action of the spin group. (One could conversely define the spin group via this property. The inverse of a product of vectors is the reverse product times $\pm1$ depending on the parity of the number of vectors, and this is extended linearly to an anti-automorphism of the algebra which is used in place of inversion for the conjugation action.) Thus, for an element $R$ of the spin group, we can define $\phi(R)$ to be a linear operator acting on the vector space via $\phi(R)v:=RvR^{-1}$.
If we assume the vector space has an ordered basis, all linear operators acting on it may be represented as matrices, which means in turn $\phi(R)$ can be written down as a matrix. This gives a $2$-to-$1$ homomorphism $\phi:\mathrm{Spin}(V)\to\mathrm{SO}(V)$. Since $\phi(-R)=\phi(R)$, the fibers are $\pm R$. In other words, $\pm R$ are distinct elements of the spin group, but turn into the same rotation in the special orthogonal group.
For illustration, in the 3D case, we can pick $\pm 1$ (where $R=1$) to be $0^\circ$ and $360^\circ$ rotations around an axis. In the vectorial world, rotating $360^\circ$ has the same final effect as rotating $0^\circ$. But in the spinorial world, we can think of an element of $\mathrm{Spin}(3)$ as a path in $\mathrm{SO}(3)$ from the identity matrix to another matrix (IOW, an "animation" of rotation), modulo endpoint-preserving homotopy. Within $\mathrm{SO}(3)$, the loop comprised of all rotations around an axis, which goes from the identity matrix back  to itself, is noncontractible. However, traversing this loop twice can be visualized as a twice-wound rubber band which actually is contractible within $\mathrm{SO}(3)$, as illustrated by Dirac's belt / plate trick.) So $\pm1$ are different elements of the spin group, but turn into the same rotation.
