numerical methods

Let's use the unknown coefficient approach to your problem. Assume that $$ (\Delta f)(x) = \frac{-f(x-2h)+16f(x-h)-30f(x)+16f(x+h)-f(x+2h)}{12h^2} $$ is a composition of two first order finite difference derivative formulas $$ \Delta f = \Delta_2(\Delta_1 f) $$ Each of the formulas has the form $$ \Delta_1 f = \frac{a_{-1} f(x-h) + a_0 f(x) + a_1 f(x+h)}{h}\\ \Delta_2 f = \frac{b_{-1} f(x-h) + b_0 f(x) + b_1 f(x+h)}{h}\\ $$ These formulas need to approximate first derivatives, so the following order conditions should hold: $$ a_{-1} + a_0 + a_1 = b_{-1} + b_0 + b_1 = 0\\ a_1 - a_{-1} = b_1 - b_{-1} = 1\\ $$ Composing these two formulas gives $$ (\Delta_2(\Delta_1 f))(x) = \frac{ (a_{-1} b_{-1}) f(x-2h) + (a_{-1} b_0 + a_0 b_{-1}) f(x-h) + (a_{-1} b_1 + a_0 b_0 + a_1 b_{-1}) f(x) + \dots }{h^2}\\ \frac{\dots + (a_0 b_1 + a_1 b_0) f(x+h) + (a_1 b_1) f(x+2h) }{h^2} $$ So we've arrived to a system of quadratic equations for $a_k, b_k$.

The problem is exactly the same as factorizing $$ p(x) = \frac{-x^4 + 16 x^3 - 30 x^2 + 16x - 1}{12} $$ into a product of $$ q(x) = a_{-1} x^2 + a_0 x + a_1\\ r(x) = b_{-1} x^2 + b_0 x + b_1 $$

Factoring the polynomial $p(x) = q(x) r(x)$ means that the roots of $p(x)$ are the union of the roots of $q(x)$ and the roots of $r(x)$ (including the multiplicity).

It is easy to see that $p(x)$ has root $x = 1$ with multiplicity 2 (this is a direct consequence of $\Delta$ being a second-order derivative approximation) and $q(x)$ and $r(x)$ also have the root $x = 1$ due to order conditions. $$ \frac{p(x)}{(x-1)^2} = \frac{x^2 - 14x + 1}{12}. $$ The polynomial in the right-hand side does not have real roots. This means that there is no factorization into $q(x) r(x)$ product with coefficients $a_k, b_k$ being real. There's no representation of the formula as a composition of two first-order three-points formulas.

Let's try some other forms of $\Delta_1$ and $\Delta_2$. $$ \Delta_1 f = \frac{a_{-1} f(x-h) + a_0 f(x)}{h}\\ \Delta_2 f = \frac{b_{-1} f(x-h) + b_0 f(x) + b_1 f(x+h) + b_2 f(x+2)}{h}\\ $$ Now $$ q(x) = a_{-1} x + a_0\\ r(x) = b_{-1} x^3 + b_0 x^2 + b_1 x + b_2 $$ The order conditions immediately give the solution for $q(x)$: $a_0 = 1, a_{-1} = -1$. Thus $\Delta_1$ is simply the left divided difference approximation. $$ (\Delta_1 f)(x) = \frac{f(x) - f(x-h)}{h}. $$ Finding $\Delta_2$ is straightforward: $$ r(x) = \frac{p(x)}{1 - x} = \frac{x^3 - 15 x^2 + 15 x - 1}{12}\\ (\Delta_2 f)(x) = \frac{f(x-h) - 15 f(x) + 15 f(x+h) - f(x+2h)}{12h}. $$ Verifying that $(\Delta_2 f)(x)$ actually approximates $f'(x)$ is left as an exercise.

Another solution may be obtained by taking $\Delta_1$ as right divided difference. It is pretty much the same solution with opposite signs and reflected nodes.

Another exercise: show that any finite difference formula of order $p$ can be represented as composition of $p-1$ order finite difference with $\frac{f(x) - f(x-h)}{h}$.