In the jj-representation, each electron from $i$ to $N$ will have:

-  $\vec{l}_i$ (orbital angular momentum), 
-  $\vec{s}_i$ (spin angular momentum), and
-  $\vec{j}_i=\vec{l}_i + \vec{s}_i$ (total angular momentum).

In your example we have $N=2$ and both electrons are $p$-type so we have:

$$\tag{1}
{l}_1=1,~~~~~~~{l}_2=1.
$$ 

Let's also assign the spins for each electron as:

$$\tag{2}
{s}_1=1/2,~~~~~~~{s}_2=1/2.
$$ 

We now have the following possibilities for ${j}_i$:

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\begin{eqnarray}
~&|l_i - s_i| \tag{3}\\
~&~~~~~~~ |l_i - s_i| +1 \tag{4}\\
~& ~\vdots \tag{5}\\
~& ~l_i + s_i. \tag{6}\\
\end{eqnarray}

Plugging in $l_i$ and $s_i$ we get the following possibilities for $j_i$:

$$\tag{7}
j_i = 1/2 , ~~~~~~~j_i=3/2.
$$

Therefore we have 4 possibilities for the $(j_1,j_2)$ pair:

$$\tag{8}
(1/2,1/2),(1/2,3/2),(3/2,1/2),(3/2,3/2).
$$

These are all included in your example. Let's now look at the total $J$ values, which come from adding up the total angular momenta $j_i$ from each of the electrons. Since we have only 2 electrons, $J$ can be:

\begin{eqnarray}
~&|j_1 - j_2| \tag{9}\\
~&~~~~~~~ |j_1 - j_2| +1 \tag{10}\\
~& ~\vdots \tag{11}\\
~& ~j_1 + j_2. \tag{12}\\
\end{eqnarray}

So the possible $J$ values in our case are (even including  some which you can eliminate due to the Pauli exclusion principle):

- for $(1/2,1/2)$: $J=0$ or $1$, which means we can have $(1/2,1/2)_0$ and $(1/2,1/2)_1$.
- for $(1/2,3/2)$: $J=1$ or $2$, which means we can have $(1/2,3/2)_1$ and $(1/2,3/2)_2$.
- for $(3/2,1/2)$: $J=1$ or $2$, which means we can have $(3/2,1/2)_1$ and $(3/2,1/2)_2$.
- for $(3/2,3/2)$: $J=0,1,2,3$, so: $(3/2,3/2)_0$, $(3/2,3/2)_1$, $(3/2,3/2)_2$, $(3/2,3/2)_3$.

For the LS-representation, we have the total orbital angular momentum:

$$\tag{13}
\vec{L}=\vec{l}_1 + \vec{l}_2,
$$

and total spin angular momentum:

$$\tag{14}
\vec{S}=\vec{s}_1 + \vec{s}_2.
$$

The possible values of $L$ are:

\begin{eqnarray}
~&|l_1 - l_2| \tag{15}\\
~&~~~~~~~ |l_1 - l_2| +1 \tag{16}\\
~& ~\vdots \tag{17}\\
~& ~l_1 + l_2, \tag{18}\\
\end{eqnarray}

and we have the same structure for $\vec{S}$.<br>
To correspond term symbols from the LS-representation with term symbols from the jj-representation, you can assign each electron an $\vec{l}_i$ and $\vec{s}_i$ and work out the corresponding $\vec{j}_i,J,L$ and $S$ values.


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This means that:

 - for $(1/2,1/2)_0$ we must have a triplet, so we get $^3P_0$ rather than $^1S_0$.
 - for $(3/2,3/2)_0$ we must have a triplet, so we get $^1P_0$ rather than $^3P_0$.
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<!-- $^1P, ^3S$ and $^3D$ are excluded due to Pauli exclusion principle -->