What is the $\frac{1}{2}$ representation of $U(1)$?
TL;DR: The label $q\in\mathbb{Q}$ (e.g. $q=\frac{1}{2}$) refers to the following (possibly multi-valued) 1-dimensional representation $$U(1)~\in~\alpha~~\stackrel{R_q}{\mapsto}~~\alpha^q ~\in~GL(1,\mathbb{C})/\sim \tag{1}$$ of the group $U(1)$.
The connected component of the (global) conformal group that contains the identity element is $$ {\rm Conf}_0(3,1)~\cong~SO^+(4,2)/ \mathbb{Z}_2~\cong~SU(2,2)/ \mathbb{Z}_4~\cong~PSU(2,2) ,\tag{2}$$ where $$ \mathbb{Z}_2~\cong~\{\pm {\bf 1}_{6\times 6} \}, \qquad\mathbb{Z}_4~\cong~\{\pm {\bf 1}_{4\times 4},\pm i{\bf 1}_{4\times 4}\},\tag{3} $$ see e.g. Ref. 2.
One can make a Lie group homomorphism $$ G~:=~SU(2)_{\ell} \times SU(2)_r \times U(1)~\ni~ (g_{\ell},g_r,\alpha)~~\stackrel{\Phi_q}{\mapsto}~~ \begin{pmatrix} \alpha^q g_{\ell} & 0 \cr 0 & \alpha^{-q} g_r \end{pmatrix}~\in~ PSU(2,2), \tag{4}$$ where $$q~\in~\frac{1}{2}\mathbb{Z}\tag{5} $$ is a half integer (or integer).
If one chooses $q=\frac{1}{2}$, one may show that the Lie group homomorphism (4) is injective. The product group $G$ is hence a subgroup of $PSU(2,2)$. This means that representations of $PSU(2,2)$ can be decomposed as representations of $G$.
Examples: In a hopefully obvious notation, the fundamental (multi-valued) representation decomposes as $$ {\bf 4}~=~ ({\bf 2}_{\ell}, {\bf 1}_r)_{\frac{1}{2}} \oplus ({\bf 1}_{\ell}, {\bf 2}_r)_{-\frac{1}{2}}, \tag{6}$$
while the complex conjugate representation decomposes as $$ \overline{\bf 4}~=~ ({\bf 2}_{\ell}, {\bf 1}_r)_{-\frac{1}{2}} \oplus ({\bf 1}_{\ell}, {\bf 2}_r)_{\frac{1}{2}}, \tag{7}$$
cf. eq. (2.1) in Ref. 1.
References:
C Romelsberger, arXiv:hep-th/0510060.
R. Penrose & W. Rindler, Spinors and Space-Time Vol. 2, 1986; p. 303-304.