I am trying to understand this statement from my Professor.
A total preorder on a set $S$ can be expressed as a total order on a partition of $S$
So for example, given $r_1 \lesssim r_2$, $r_2 \lesssim r_1$, $r_2 \lesssim r_3$ and $r_3 \not \lesssim r_2$ is equivalent to $\{ r_1, r_2\} < \{r3 \}$
Am I right?
You're right. A total preorder is not a total order only because different elements can be “equal” under it, i.e. $a\le b$ and $b\le a$ doesn’t imply $a=b$.
To show that a total preorder induces a total order on a partition of $S$, consider the relation
$$r\sim s\Leftrightarrow r\le s \land s\le r\;.$$
This is an equivalence relation – it’s clearly symmetric, and it inherits reflexivity and transitivity from the preorder. It partitions $S$ into equivalence classes of elements that are “equal” under the preorder.
The preorder induces a total order on the equivalence classes of $S$ under this equivalence relation:
$$[r]\preccurlyeq[s]\Leftrightarrow r\le s\;.$$
This is well-defined since it doesn’t depend on the representatives: If $r\le s$, $r\sim t$ and $s\sim u$, then $t\le r\le s\le u$ and thus by transitivity $t\le u$. It’s a total order because it inherits reflexivity, transitivity and totality from the preorder, and it’s antisymmetric: If $[r]\preccurlyeq[s]$ and $[s]\preccurlyeq[r]$, then by the definition of $\preccurlyeq$ we have $r\le s$ and $s\le r$, and then by the definition of $\sim$ we have $r\sim s$ and thus $[r]=[s]$.
I’ll leave the details of the “inheritance” claims as an exercise.
Conversely, every total order $\preccurlyeq$ on a partition of $S$ induces a total preorder $\le$ on $S$, again via
$$ r\le s\Leftrightarrow [r]\preccurlyeq[s]\;. $$
Reflexivity, transitivity and totality are inherited from $\preccurlyeq$, but antisymmetry is not (since $[r]=[s]$ doesn’t imply $r=s$).
These two transformations are inverses of each other, so the total preorders on $S$ and the total orders on partitions of $S$ are in one-to-one correspondence.