By the Sylow counting theorems, the number $n_2$ of Sylow 2-subgroups is congruent to 1 mod 2 and divides $24/2^3=3$. Thus, $n_2$ equals 1 or 3. If $n_2$ equals 1, then $D_4$ is the unique Sylow-2 subgroup in $S_4$, which means it is also normal in $S_4$, but this is not the case. For it can be seen that $D_4$ is not closed under conjugates: it contains $(1234)$ but not $(1324)$, for example, or it contains $(13)$ but not $(12)$. Thus $n_2=3$. One of the subgroups in $S_4$ of order 8 is $D_4 = \langle (13), (1234) \rangle$. By the Sylow theorems, all subgroups of order 8 are conjugate. Hence the other two subgroups of order 8 can be obtained by relabeling the points in $D_4$ (i.e. taking conjugates of $D_4$). For example, conjugation by $(23)$ gives the subgroup $\langle (12), (1324) \rangle$, and conjugation by $(34)$ gives the subgroup $\langle (14), (1243) \rangle$.
Regarding your question on how to choose $\sigma$, it is known that the $n$-cycle $(1,2,\ldots,n)$ and the transposition $(i,j)$ together generate all of $S_n$ iff $i-j$ and $n$ are relatively prime.
Assume $f$ acts as identity on $P_i$ and as inversion on $P_j$. Let $d$ be the point fixed by $P_i$ and $c$ the point fixed by $P_j$, and $a,b$ the other two. Then
$$f((a\,c)(b\,d))=f(\underbrace{(a\,b\,c)}_{\in P_i}\underbrace{(a\,b\,d)}_{\in P_j})= (a\,b\,c)(a\,d\,b)= (a\,d\,c),$$
which is impossible (order 2 $\ne$ order 3).
And if $f$ acts as inversion on both $P_i$ and $P_j$,
$$ f((c\,b\,d))=f((a\,c\,b)(a\,b\,d))=(a\,b\,c)(a\,d\,b)=(a\,d\,c),$$
so $f$ maps $\langle (b\,c\,d)\rangle$ to $\langle (a\,c\,d)\rangle$, contradiction.
We conclude that the only possibility is that $f$ acts as identity on all $P_i$.
Best Answer
Not sure how to "convince" you, but I suspect you are worried because some of the 2-cycles do the job and others do not. The trick is in picking the right 4-cycle to go with your 2-cycle or vice versa. (1,2) works with (1,2,3,4) because 1 and 2 are adjacent in (1,2,3,4) whereas 1,3 is not. But (1,3) will work with (1,3,2,4).
For more info:
http://en.wikipedia.org/wiki/Symmetric_group#Generators_and_relations In particular it says the following is a generating set: "a set containing any n-cycle and a 2-cycle of adjacent elements in the n-cycle."