$u(x+h)-u(x)=c p(x') h$ for some $x'$ between $x, x+h$, what derivative rule is this?
$c$ is constant, $p$ is continuous function, $x,h \in \mathbb{R}$. Dividing by $h$ and taking limit $\rightarrow \infty$ one gets:
$$u'(x)=cp(x)$$
But I forgot what this idea about "for some $x'$ between $x,x+h$" is?
For some $u(x)$ continuous in $[x,x+h]$ and differentiable over $(x,x+h)$, by the Mean Value Theorem :
$$\exists x_0 \in (x,x+h) : u(x_0) = \frac{u(x+h)-u(x)}{h}$$
Now, for whatever reason in the statement you found, it seems that it is something like
$$u(x_0)\cdot h = u(x+h)-u(x) \implies chp(x_0) = u(x+h)-u(x)$$
where a function $p(x)$ such that $u(x_0) = cp(x_0)$ is used.
Without any further context though, it's hard to understand the intuition behind this.