First of all it should be noted that if we prove the assertion for cycles only then we're done. Indeed if the permutation $p$ has cycle decomposistion $p = c_1c_2c_3\ldots c_n$ and each cycle $c_i = r_is_i$ with $r_i, s_i$ of order $2$ then $r_i$ and $s_i$ commute with $r_j$ and $s_j$ with $i \neq j$ since they act on different points (like the $c_i$ do). So we can write $p = r_1s_1r_2s_2 \ldots r_ns_n$.In a first step we can move $s_1$ to the end giving $p = r_1r_2s_2\ldots r_ns_1s_n$. In a second step we can move $s_2$ to behind $s_1$ giving $p = r_1r_2r_3s_3\ldots r_ns_1s_2s_n$. In a finite number of analoguous steps we end up with $p = r_1r_2\ldots r_ns_1s_2\ldots s_n$. where $r_1r_2\ldots r_n$ and $s_1\ldots s_n$ have order $2$.
Now let $p = (i_1, i_2, \ldots,i_n)$ be a cycle of length $n$. I will call a rotation a move that can be compared to the movement of an airplane propellor, e.g. for a tuple $[1,2,3,4,5]$ the rotation will be $[5,4,3,5,1]$ this rotation is realized by the permutation $r = (1,5)(2,4)$, leving the point $3$ fixed. Let $s$ be the rotation on the tuple with the first element discarded, in our examploe $[2,3,4,5]$ giving $[5,4,3,2]$ realized by the rotation $s = (2,5)(3,4)$. We can now realize visually that starting from a given point and applying a rotation $r$ and then a rotation $s$ then the points ends up with its successor. If necessary write the numbers on a strip of paper put a needle in the midpoint of the rotation, do $r$ replace the needle to the midpoint of the rotation $s$ and apply $s$ a convince yourself that you recovered the permutation as $p = rs$ with $r$ and $s$ of order $2$.