We consider a 'neat' function $f$ near a point $x_0$. The second derivative $f''$ is of course the derivative of the first derivative $f'$. If you apply the 3-point central difference formula repeatedly, you can therefore find an approximation formula for the second derivative in $x_0$ at a step size $h$.

$$\begin{aligned} f''(x_0)&\approx \frac{f'(x_0+h)-f'(x_0-h)}{2h} \\ \\ &\approx\frac{\displaystyle\frac{f(x_0+2h)-f(x_0)}{2h}-\frac{f(x_0)-f(x_0-2h)}{2h}}{2h}\\ \\ &=\frac{f(x_0+2h)-2f(x_0)+f(x_0-2h)}{(2h)^2}\end{aligned}$$

But now you see that you need the function values in $x_0-2h$, $x_0$ and $x_0+2h$. However, if you use imaginary points on $x_0-\tfrac{1}{2}h$ and $x_0+\tfrac{1}{2}h$, you end up with the following 3-point central difference formula for the second derivative:

A more formal derivation of the 3-point central difference formula for the second derivative provides more insight into the truncation error.

We look at two Taylor approximations around $x_0$:

$$\begin{aligned} f(x_0+h)=f(x_0)+f'(x_0)h+\tfrac{1}{2}f''(x_0)h^2+\tfrac{1}{6}f'''(x_0)h^3+ O(h^4)\\[3pt] f(x_0-h)=f(x_0)-f'(x_0)h+\tfrac{1}{2}f''(x_0)h^2-\tfrac{1}{6}f'''(x_0)h^3+ O(h^4)\end{aligned}$$ When we add the two equations, we get: $$f(x_0+h)+ f(x_0-h)=2f(x_0)h +f''(x_0)h^2 +O(h^4)$$ Thus: $$\frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2}=f''(x_0)+ O(h^2)$$ The truncation error of this 3-point central difference approximation is quadratic in $h$.