Second Order Derivatives in Becker's Crime and Punishment
You have to apply both chain and product rule for derivatives here. In this case:
$$C_p'=\frac{\partial C(pO)}{\partial p}=C_p'(pO)O$$
But note when calculating cross-derivative you take the 2nd order derivative wrt O:
$$\frac{\partial C_p'(pO)O}{\partial O} =  \frac{\partial }{\partial O}[C_p'(pO)]\cdot O +  C_p'(pO)  \cdot\frac{\partial}{\partial O} [O] \\
= C_{pO}''(pO) \cdot pO + C_p'(pO)  \cdot 1 $$
Omitting subscripts this is equivalent to: $C'' pO + C'$.
To sum up, $C'$ comes from the product rule for derivatives (i.e. $\frac{d}{dx}[f(x)g(x)]= f'(x)g(x)+ f(x)g'(x)$ - you can learn more about this rule in standard math textbooks such as EMEA by Hammond et al.).