I'll use a longer cycle to help describe two techniques for writing disjoint cycles as the product of transpositions:
Let's say $\tau = (1, 3, 4, 6, 7, 9) \in S_9$
Then, note the patterns:
Method 1: $\tau = (1, 3, 4, 6, 7, 9) = (1, 9)(1, 7)(1, 6)(1, 4)(1, 3)$
Method 2: $\tau = (1, 3, 4, 6, 7, 9) = (1, 3)(3, 4)(4, 6)(6, 7)(7, 9)$
Both products of transpositions, method $1$ or method $2$, represent the same permutation, $\tau$. Note that the order of the disjoint cycle $\tau$ is $6$, but in both expressions of $\tau$ as the product of transpositions, $\tau$ has $5$ (odd number of) transpositions. Hence $\tau$ is an odd permutation.
Now, don't forget to multiply the transpositions you obtain for each disjoint cycle so you obtain an expression of the permutation $S_{11}$ as the product of the product of transpositions, and determine whether it is odd or even:
$\sigma = (1, 4, 10)(3, 9, 8, 7, 11)(5, 6)$.
If you decompose into cycles first, all you need to do is express each cycle as a product of transpositions. There are various ways to do this, for example
$$ (1\,2\,3\,4\,\ldots\,n) = (1\,n)\cdots(1\,4)(1\,3)(1\,2) $$
or
$$ (1\,2\,3\,4\,\ldots\,n) = (1\,2)(2\,3)(3\,4)\cdots(n{-}1\;n) $$
Best Answer
In product notation the permutations are normally applied from right to left. Be sure to distinguish the product of two permutations from the concatenation of cycles within a single permutation, by using a dot ($\cdot$) to signify 'product'.
Verify the second example on a string like $abcd$: $$ abcd \stackrel{(243)}\to acdb\stackrel{(1243)}\to dabc $$ is the same as: $$ abcd\stackrel{(14)}\to dbca\stackrel{(34)}\to dbac\stackrel{(23)}\to dabc. $$ This proves that $(1243)\cdot(243)=(23)\cdot(34)\cdot(14)$.
You have a typo in your first example. The assertion $(132)=(13)\cdot(12)$ is false, since $$ abc\stackrel{(132)}\to bca $$ while $$ abc\stackrel{(12)}\to bac\stackrel{(13)}\to cab $$ But it is true that $(132)=(12)\cdot(13)$: $$ abc\stackrel{(13)}\to cba\stackrel{(12)}\to bca.$$
The product of a transposition with itself is the identity. The transposition $(ij)$ swaps element $i$ with element $j$. Doing this a second time will return the elements to their original places.