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:
$\sigma = (1, 4, 10)(3, 9, 8, 7, 11)(5, 6)$.
Once a permutation is known as a product of disjoint (hence commuting) cycles, one can work in parallel and so it suffices to know how to write a cycle as a product of transpositions. To avoid a clutter of subscripts, I'll give an example:
$(7413625) = (75)(72)(76)(73)(71)(74)$
(In each of these transpositions 7 occurs, the other numbers accompanying 7 are taken in the reverse order to the given cycle)
Best Answer
As for the disjoint cycle, first pick any element in $\{1,2,3,4,5,6\}$. For simplicity, we'll pick $1$. So we'll start our cycle by writing, $$(1$$
Now, we see what element $1$ gets sent to. We look for $1$ in the top row and see that $2$ is the corresponding element in the bottom row, so $1 \to 2$, so continuing our cycle, $$(12$$
Next, we look for $2$ in the top row and see that $4$ is the corresponding element in the bottom row, so $2 \to 4$, thus our cycle so far is, $$(124$$
Then just keeping continuing in this fashion until we run into an element we've already seen. Notice that $4 \to 5, 5 \to 6, 6 \to 3$ so our cycle is, $$(124563$$
Since $3 \to 1$ and $1$ is already in the cycle, that is our cue to end the cycle at $3$, $$(124563)$$
Notice that we've exhausted all the elements in the permutation, so we're done. If we weren't done, we'd just pick an element not yet in any of our cycles, and start again using the same process.
As for transpositions, notice that each transposition is a two-cycle, or in other words, a "swap". So you can think of writing a product of transpositions as the sequence of swaps you need to make to rearrange the numbers $1, 2, 3, 4, 5, 6$ as $2, 4, 1, 5, 6, 3$.