Why do we have residual symmetry when we already used all symmetry in gauge fixing of the worldsheet metric? In Becker, Becker and Schwarz' book about string theory, the following symmetries are listed for the $\sigma$-model of the string:

*

*Poincaré transformations

*Reparametrizations $\sigma^{\alpha}\rightarrow f^{\alpha}(\sigma)$ and corresponding pullback of the worldsheet metric $h_{\alpha\beta}$

*Weyl transformations $h_{\alpha\beta}\rightarrow e^{\phi(\sigma,\tau)}h_{\alpha\beta}$
To get to $h_{\alpha\beta}=\mathrm{diag}(-1,1)$ gauge we use the following: As $h_{\alpha\beta}$ is symmetric it only has $3$ independent parameters. By reparametrization invariance, we get two equations which reduces the number of independent parameters to 1 and the last one is removed by using Weyl transformation symmetry.
But what bugs me is that even after this there is a symmetry left that BBS term residual symmetry, which is related to reparameterization symmetry: A reparameterization satisfying
$$\partial_{\alpha}\xi_{\beta}+\partial_{\beta}\xi_{\alpha}=\Lambda \eta_{\alpha\beta}$$
is still a symmetry after gauge fixing.
Since we already "used up" Weyl and reparametrization symmetries, why is this residual symmetry left?
 A: The whole reason behind gauge fixing is to make the path integrals easier: choosing a slice that cuts each gauge equivalence class once allows us to get rid of the gauge volume via the Fadeev-Popov procedure. While we're at it, we might as well gauge fix the metric to the 2D unit metric (at least locally), since it's nice and simple. However, we could have made any other choice of gauge fixing, provided that it is related to the unit metric via diffeomorphisms and Weyl scalings. It turns out that in 2D, all metrics are related via these symmetries (proof).
Now once we've fixed the metric to the unit metric, we would expect any infinitesimal transformation of $\mathrm{Diff}\times\mathrm{Weyl}$ to push the metric away from our gauge choice. However, there are transformations that don't do that: these are precisely the diffeomorphisms that can be undone by Weyl scalings, i.e. the conformal transformations. This is the "residual symmetry", but it's not "residual" in the sense that it's not covered by $\mathrm{Diff}$ or $\mathrm{Weyl}$. It's "residual" in that it remains even after gauge fixing. All of this stems from the fact that string theory is formulated with a dynamical background metric, which means that conformal transformations are honest diffeomorphisms.
However, we have seen that we kill off 3 degrees of freedom via our gauge transformations, corresponding to the 3 free components of a symmetric metric (or antisymmetric, if we were in Minkowski space). So doesn't the presence of residual conformal symmetry contradict the original statement? No, it's consistent because $\mathrm{Conf}\subset\mathrm{Diff}$ is a measure-zero subset and so sneaks past the counting procedure. As a side point, the presence of the residual symmetry (manifesting as "conformal Killing vectors") is not as annoying as it seems, much of it is handled automatically by boundary conditions.
This is probably easier to illustrate like this: imagine our theory has a symmetry under a group $G$ with an action on the space of metrics $G\times H\to H$. We could gauge fix our metric to some $h\in H$ of our choice using the action of $G$. However, we might notice that there are elements $g\in G$ which act as $gh=h$ on our choice of metric. So just because we have "used up" the symmetry $G$ in gauge fixing does not mean that there can't be residual symmetry. This is precisely what happens for the string worldsheet, except for the fact that the symmetry group is roughly a product of $G=\mathrm{Diff}\times\mathrm{Weyl}$.
Finally, if you want to derive the infinitesimal form of these residual transformations:
$$
g'_{\mu\nu}(x')=\Omega^2(x)g_{\mu\nu}(x)
\\g'_{\rho\sigma}(x')\frac{\partial x'^\rho}{\partial x^\mu}\frac{\partial x'^\sigma}{\partial x^\nu}=\Omega^2(x)g_{\mu\nu}(x)
\\x'_\mu=x_\mu+\epsilon_\mu+\mathcal O(\epsilon^2)
\\g'_{\rho\sigma}(x')(\delta^\rho_\mu+\partial_\mu\epsilon^\rho)(\delta^\sigma_\nu+\partial_\nu\epsilon^\sigma)=\Omega^2(x)g_{\mu\nu}(x)
\\g_{\mu\nu}+(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)+\mathcal O(\epsilon^2)=\Omega^2(x) g_{\mu\nu}
\\\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\kappa(x)g_{\mu\nu}\qquad\square
$$
