$f(x)= (x+1)^4 $
points where function is measured/sampled are 0 and 1. I thought when we have two conservative derivative on same point$\ x$ as $f'(x)$, in next column their divided difference becomes $f''(x)$
so ,in the image, why then we have $ \frac{1}{2}f''(0)$ instead of just $f''(0)$
Try to construct this term in the normal way. It is easy to show that
$$f[x,x+h,x+2h]=\frac{f(x+2h)-2f(x+h)+f(x)}{2h^2}.$$
If $f$ is twice differentiable at $x$, then this difference tends to $\frac{1}{2}f^{''}(x),$ whenever $h\to 0$.