Algebra of Noether's charges and algebra of symmetry transformations

I’m trying to understand the connection of algebra of transformations under a commutator and algebra of Noether’s charges under Poisson bracket.

I have a problem that results I infer from theoretical consideration do not match with results I get for simple example – free scalar theory in two dimensions. I am making a mistake somewhere but I am not sure where.

Theoretical approach:

Let us assume we have symmetry transformation $f^mu$ such that
$$delta_f phi(x) = phi'(x) – phi(x) = f^mu(x)partial_muphi(x).$$

From this I would expect that commutator of the symmetry transformations is
$$[delta_f, delta_g]phi = (f^mupartial_mu g^nu – g^mupartial_mu f^nu)partial_nuphi equiv h^nupartial_nuphi = delta_hphi.$$

There are a lot questions on this page proving and discussing that conserved Noether’s charge $Q_f$ generates symmetry, i.e.
$$delta_fphi(x) = {phi(x), Q_f},$$
where the Poisson bracket is
$${A(textbf{x},t), B(textbf{y},t)} =int d{textbf{x’}} frac{delta A(textbf{x},t)}{deltaphi(textbf{x’},t)}frac{delta B(textbf{y},t)}{deltapi(textbf{x’},t)} – frac{delta A(textbf{x},t)}{deltapi(textbf{x’},t)}frac{delta B(textbf{y},t)}{deltaphi(textbf{x’},t)}$$

From here I would expect

$$delta_hphi= [delta_f,delta_g]phi = delta_fdelta_gphi – delta_gdelta_fphi = {{phi, Q_g},Q_f} – {{phi,Q_f},Q_g} $$
$$stackrel{text{Jacobi id.}}{=} -{{Q_g, Q_f}, phi} = -{phi,{Q_f,Q_g}} equiv {phi,Q_h}$$

From which I conclude
$$Q_h = -{Q_f, Q_g}$$

Example of free scalar field:

I want to illustrate previous considerations on a simple theory

$$ I = int d^2x frac12 partial_muphipartial^muphi$$

This action is invariant under $f^mu$ being conformal Killing vector in flat space, i.e. $f^mu$ satisfy
$$partial_mu f_nu + partial_nu f_mu = eta_{munu}partial_rho f^rho.$$

Using Noether’s theorem I derived the conserved charge associated with the symmetry
$$Q_f = int dx Big[frac12((partial_xphi)^2+pi^2)f^t+f^xpipartial_xphiBig]$$

From this I have
$$frac{delta Q_f}{deltapi(x,t)} = f^tpi + f^xpartial_xphi, qquad frac{delta Q_f}{delta phi(x,t)} =-partial_x( f^tpartial_xphi+pi f^x)$$

This charge really satisfies equation
$$delta_f = {phi, Q_f}$$

But the problematic result I get is
$${Q_f, Q_g} = Q_h.$$
No minus is here.

I believe the ambiguity is in evaluating
$$delta_fdelta_gphi,$$
because Poisson bracket ignores the $g^mu$ and $f^mu$. While in the theoretical approach I expected
$$delta_f(g^mupartial_muphi)= f^nupartial_nu(f^mupartial_muphi)$$
there I didn’t ignore $g^mu$.

But I am missing the physical reasoning here and I do see know which result is correct.