Approximating the 1st order derivative via central differences can be written as $ \delta_{2h}u(x) =\frac{u(x+h) - u(x-h)}{2h} \approx u'(x) .$
What is the main issue with applying again a central difference to compute $u''(x)$?
In this case,
$\delta_{2h}u'(x) = \frac{u'(x+h) - u'(x-h)}{2h} \approx \frac{u(x+2h) + u(x-2h) - 2u(h)}{4h^2}.$
This is actually different from what most sources on finite differences consider the second order approximation using central differences, i.e.
$ u''(x) \approx \frac{u(x+h)+u(x-h)-2u(x)}{h^2},$
which is achieved for a step-size of $\frac{1}{2}h$.
What is the problem with the first interpretation and why do most sources require that one uses the half step size (apart from the obvious loss of numerical accuracy, is there any logical fallacy)?
$$u(x+h)= u(x)+u'(x)h+\frac{1}{2}u''(x)h^2+\mathcal{O}(h^2)$$ $$u(x-h)= u(x)-u'(x)h+\frac{1}{2}u''(x)h^2+\mathcal{O}(h^2)$$
The common formula is obtained not assuming "half step size" , but by adding the above formula together.
Your result is similar to looking at the below formula, the steps are larger. $$u(x+2h)= u(x)+u'(x)2h+\frac{1}{2}u''(x)(2h)^2+\mathcal{O}((2h)^2)$$ $$u(x-2h)= u(x)-u'(x)2h+\frac{1}{2}u''(x)(2h)^2+\mathcal{O}((2h)^2)$$