This theorem has either been mis-interpreted by the question author or there is a mistake in the referenced book. Consider the following counter example:
$$f(x) = 100+x$$ $$h=0.01$$ $$R = 0.01$$
At $x=0$ the absolute error in each function evaluation is $100*0.01=1$, so we have
$$f'^-(0) = \frac{f^-(h) - f^- (-h)}{2h} = \frac{(100+0.01)\pm 1 - ((100-0.01) \pm 1)}{0.02} $$
$$ \implies f'^-(0) = \frac{0.02}{0.02} \pm \frac{1}{0.02}\pm \frac{1}{0.02}$$
In the worst case scenario, the two error terms have the same sign and do not cancel out. The relative error of the derivative approximation can therefore be as much as $100$, which is much greater than $R/h = 1$.
As far as I can tell there is no bound on the relative error for a general $f$ since by choosing a function of the form $f(x) = n+x$, the relative error in the derivative approximation can always be increased simply by increasing $n$.
On the other hand we can calculate a bound dependent on $f$. The bound on the absolute error for sufficiently small $h$ and $R$ is:
$$ f'^-(x) = \frac{f(x+h)-f(x-h)}{2h} \pm \frac{f(x)R}{h} $$
Proof:
$$f^-(x+h) = (f(x)+hf'(x))(1 \pm R)$$
$$ \implies f^-(x+h) =f(x)+hf'(x)\pm Rf(x)$$
where we Taylor expand $f$ around $x$ and neglect terms of order $hR$ or $h^2$ or higher since both $h$ and $R$ are small. Similarly
$$ f^-(x-h) = f(x) - hf'(x) \pm Rf(x)$$
Therefore
$$f'^-(x) = \frac{2hf'(x) \pm Rf(x) \pm Rf(x)}{2h} $$
$$ \implies f'^-(x) =f'(x) \pm \frac{Rf(x)}{h}$$
where we once again consider the worst case scenario, in which the errors add up.
The bound on the relative error is therefore dependent on $f(x)$ and can be expressed as
$$ f'^-(x) = f'(x)\left(1\pm\frac{Rf(x)}{hf'(x)}\right) $$
Similarly we have
$$f''^-(x) = f''(x)\left(1 \pm \frac{4Rf(x)}{h^2f''(x)}\right)$$
