How do I arrive at the multiplication function in lambda calculus?
You know that $(\bar{n}\ s)$ corresponds to $s^{(n)}$, i.e. function $s$ applied $n$ times to its argument. You want to obtain a function that iterates some function $s$ exactly $n \cdot m$ times. From the above observations and some simple rules for the composition of functions we can immediately derive the body of $\textsf{mult}$:

$$s ^ {(n \cdot m)} =
\left( s ^ {(m)} \right) ^ {(n)} = 
\bar{n} \left(s ^ {(m)} \right) = 
\bar{n}\ (\bar{m}\ s) 
$$

Hence

$$ \textsf{mult}\ :=\ \overbrace{\lambda n m.}^\text{  parameters}\underbrace{(\lambda sz.n\ (m\ s) z)}_{\text{Church numeral}}$$

or, if we $\eta$-contract, we get

$$ \textsf{mult}\ :=\ \lambda nms.n\ (m\ s)$$