Figures are a log easier to look at then a table when analyzing finite automata's. I have taken the liberty of constructing the automata given into something more recognizable:

Where state 0 and 6 are start and accepting states, respectively and all unlabeled transitions go back to state 0.
My version of Cormen, Leiserson and Rivest is older than yours (the algorithm you reference is on page 868 in mine) but I believe the content is the same. Here is the algorithm they present:
$\text{Compute-Transition-Function}(P, \Sigma)$
1 $\ \ \ \ m \leftarrow length[P] $
2$\ \ \ \ {\bf \text{for } } q \leftarrow 0 {\bf \text{ to }} m $
3$\ \ \ \ \ \ \ \ {\bf \text{do for }} \text{each character } a \in \Sigma $
4$\ \ \ \ \ \ \ \ \ \ \ \ {\bf \text{do }} k \leftarrow \text{min}(m+1, q+2)$
5$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\bf \text{repeat }} k \leftarrow k - 1 $
6$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\bf \text{until }} P_k \sqsupset P_q a $
7$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta(q,a) \leftarrow k $
8$\ \ \ \ {\bf \text{return }} \delta $
They want to construct a DFA of length one longer than the pattern, $P$. The state transition is stored in $\delta$. $\delta(q,a) \leftarrow k$ adds an edge with label $a$ from state $q$ to state $k$. They loop through each state and for each state loop through each character in the alphabet to determine where that state should go.
I believe the problem you are having is with the suffix function on line 6, where the suffix function is '$\sqsupset$'.
The suffix function is defined as follows:
$ w \sqsupset x \rightarrow \exists y \ \text{ s.t. } x = yw $
meaning, $w$ can be found as a suffix of $x$.
In words, the loop that line 6 is contained in is saying "Find the maximum suffix of the strings shorter than the current state $q$ that start at $P[0]$". Once that is found, we know we can 'skip ahead' and transition to that state instead of the beginning. There are some fence post issues and that's why they have the min function there, but the intuition still holds.
By considering all strings, $P_k$, that start at the beginning of the pattern shorter than the current position of the read pattern, $P_q a$, we test each $P_k$ to be a suffix of $P_q a$ and choose the maximum. This is just a more formal and generalized way of expressing your intuition.