As noted in Vit's post, the diagonal group $G_D \subset G \times G$ consists of the elements $\{(g,g) : g \in G \}$ with multiplication defined to be $(g,g) \cdot (h,h) = (gh,gh)$. A representation $(R_1,R_2)$ of $G \times G$ thus transforms as the representation $R_1 \otimes R_2$ under $G_D$, where $\otimes$ denotes the tensor product.

Let me illustrate through an example in the paper you are reading. In equation (2.16), it is stated that the fermions are in the representation
\begin{equation}
(1/2,0,1/2)^1 \oplus (0,1/2,1/2)^{-1}
\end{equation}
under $SU(2)_L\times SU(2)_R \times SU(2)_I \times U(1)$. Now under
\begin{equation}
SU(2)_L\times SU(2)_{R'} \times U(1) \subset SU(2)_L\times SU(2)_R \times SU(2)_I \times U(1) \,,
\end{equation}
where $SU(2)_{R'}$ is the diagonal subgroup of $SU(2)_R \times SU(2)_I$, this representation transforms as
\begin{equation}
(1/2,0 \otimes 1/2)^1 \oplus (0,1/2 \otimes 1/2)^{-1} \,.
\end{equation}
We know the tensor product decomposition rule for $SU(2)$ pretty well:
\begin{equation}
0 \otimes 1/2 = 1/2 \,, \quad
1/2 \otimes 1/2 = 0 \oplus 1 \,.
\end{equation}
Recall that the numbers here denote the "spin" of the representation. A spin $j$ representation has dimension $(2j+1)$. Note that the representation ``0" is the trivial (one-dimensional) representation. Thus we recover equation (2.18):
\begin{equation}
(1/2,1/2)^1 \oplus (0,0)^{-1} \oplus (0,1)^{-1} \,.
\end{equation}

I hope this makes everything clear to you now.