Start by writing out $R$ explicitly:
$$R=\{\langle 0,1\rangle,\langle 1,4\rangle,\langle 2,7\rangle,\langle 3,10\rangle,\langle 4,13\rangle,\langle 5,16\rangle,\langle 6,19\rangle,\langle 7,22\rangle,\langle 8,16\rangle,\langle 8,25\rangle,\langle 9,28\rangle\}$$
Now look for the ‘linked’ pairs, like $\langle 0,\color{brown}1\rangle$ and $\langle\color{brown}1,4\rangle$: transitivity says that when you have linked pairs like that in the relation, you must also have corresponding the ‘shortcut’ pair, in this case $\langle 1,4\rangle$. Here the linked pairs are:
$$\begin{align*}
&\langle 0,1\rangle\quad\text{and}\quad\langle 1,4\rangle\;,\\
&\langle 1,4\rangle\quad\text{and}\quad\langle 4,13\rangle\;,\text{ and}\\
&\langle 2,7\rangle\quad\text{and}\quad\langle 7,22\rangle\;,
\end{align*}$$
so we have to add the shortcut pairs $\langle 0,4\rangle$, $\langle 1,13\rangle$, and $\langle 2,22\rangle$.
Now repeat the process: for example, we now have the linked pairs $\langle 0,4\rangle$ and $\langle 4,13\rangle$, so we need to add $\langle 0,13\rangle$. When you finish a second pass, repeat the process again, if necessary, and keep repeating it until you have no linked pairs without their corresponding shortcut.
If $R,S$ are relations, then define $RS$ (or $R \circ S$) as
$R \circ S = \{ (x,z) | \exists y \ xRy \text{ and } y S z\}$.
Let $R^k$ be the composition of $R$ with itself $k$ times.
Then define $R^* = \cup_{k=1}^\infty R^k$. Since $R$ above is finite, then one can explicitly compute $R^*$ in a finite number of steps by a 'fixed point'
computation starting with $R$.
By 'fixed point', I mean to find the smallest set $Q$ such that $R \subset Q$ and $RQ \subset Q$, and the scheme is $Q_0 = R$, $Q_{k+1}= Q_k \cup R Q_k$, stopping when $Q_{k+1} = Q_k$.
Best Answer
I suspect that "Using the connectivity relation" to find the transitive closure is the following algorithm. Start with $T=\emptyset$ as a beggining of the transitive closure
Note that after we have done up to step 5. for $a$ T will consist of $\{(a,a),(a,b),(a,c),(a,d),(a,e)\}$
This method is essentially to write the graph which R forms, and see which elements you may reach from each node, and add these edges to the graph, why you may call it the connectivity relation.