It seems your textbook is using the right-to-left multiplication convention, so that you read the permutations from the right to the left much like you would with any other composition of maps.
As Lord Shark the Unknown has written, it is correct that
$$
(1 \quad 2 \quad 3) = (2 \quad 3 \quad 1),
$$
as these are both the cycles that map
$$
1 \mapsto 2, \quad 2 \mapsto 3, \quad 3 \mapsto 1.
$$
The third equivalent cycle is $(3 \quad 1 \quad 2)$.
Now in the multiple
$$
(1 \quad 2)(2 \quad 3),
$$
we read the cycles from left to right and make a note of where each element is mapped. Let us write $\sigma = (1 \quad 2)$ and $\gamma = (2 \quad 3)$, so we are finding the multiple $\sigma\gamma$,
$$
\stackrel{\sigma}{(1 \quad 2)}\stackrel{\gamma}{(2 \quad 3)}.
$$
We also note that since $\gamma = (2 \quad 3)$ has no mention of $1$, it must fix $1$, so that $1 \mapsto 1$ under $\gamma$, and similarly $3 \mapsto 3$ under $\sigma$. We observe that
$$
1 \stackrel{\gamma}{\mapsto} 1 \stackrel{\sigma}{\mapsto} 2 \implies 1 \stackrel{\sigma\gamma}{\mapsto} 2.
$$
Similarly
$$
2 \stackrel{\gamma}{\mapsto} 3 \stackrel{\sigma}{\mapsto} 3 \implies 2 \stackrel{\sigma\gamma}{\mapsto} 3, \quad 3 \stackrel{\gamma}{\mapsto} 2 \stackrel{\sigma}{\mapsto} 1 \implies 3 \stackrel{\sigma\gamma}{\mapsto} 1.
$$
Hence
$$
1 \stackrel{\sigma\gamma}{\mapsto} 2, \quad 2 \stackrel{\sigma\gamma}{\mapsto} 3, \quad 3 \stackrel{\sigma\gamma}{\mapsto} 1
$$
is written in the cycle notation as
$$
\sigma\gamma = (1 \quad 2 \quad 3).
$$
I will leave it to you to confirm that
$$
(2 \quad 3)(1 \quad 2) = (1 \quad 3 \quad 2).
$$
As for larger compositions, we apply the same method. For example, in the product
$$
(1 \quad 2)(2 \quad 3)(3 \quad 4)
$$
you can do it in parts the way that you have, but a (maybe simpler) way to do it is the following. Let us write
$$
\sigma = (1 \quad 2), \quad \gamma = (2 \quad 3), \quad \tau = (3 \quad 4)
$$
so that we are finding the product $\sigma\gamma\tau$,
$$
\stackrel{\sigma}{(1 \quad 2)}\stackrel{\gamma}{(2 \quad 3)}\stackrel{\tau}{(3 \quad 4)}.
$$
Also remember that, for example, since both $3$ and $4$ are not mentioned in $\sigma = (1 \quad 2)$, we have that $3 \stackrel{\sigma}{\mapsto} 3$ and $4 \stackrel{\sigma}{\mapsto} 4$. We evaluate the product reading from right to left and check where each element goes, so that
$$
1 \stackrel{\tau}{\mapsto} 1 \stackrel{\gamma}{\mapsto} 1 \stackrel{\sigma}{\mapsto} 2 \implies 1 \stackrel{\sigma\gamma\tau}{\mapsto} 2.
$$
Similarly
$$
\begin{array}{c}
2 \stackrel{\tau}{\mapsto} 2 \stackrel{\gamma}{\mapsto} 3 \stackrel{\sigma}{\mapsto} 3 \implies 2 \stackrel{\sigma\gamma\tau}{\mapsto} 3 \\
3 \stackrel{\tau}{\mapsto} 4 \stackrel{\gamma}{\mapsto} 4 \stackrel{\sigma}{\mapsto} 4 \implies 3 \stackrel{\sigma\gamma\tau}{\mapsto} 4 \\
4 \stackrel{\tau}{\mapsto} 3 \stackrel{\gamma}{\mapsto} 2 \stackrel{\sigma}{\mapsto} 1 \implies 4 \stackrel{\sigma\gamma\tau}{\mapsto} 1.
\end{array}
$$
Thus
$$
\sigma\gamma\tau = (1 \quad 2)(2 \quad 3)(3 \quad 4) = (1 \quad 2 \quad 3 \quad 4).
$$
You can similarly show that
$$
\tau\gamma\sigma = (3 \quad 4)(2 \quad 3)(1 \quad 2) = (1 \quad 4 \quad 3 \quad 2).
$$
