Let $(M,g)$ be a Riemannian manifold and consider a classical field theory $\phi: M \to \mathbb{R}$ given by some action functional $S$ satisfying Weyl & diffeomorphism invariance $$ S(\phi, e^\omega g) = S(\phi,g), $$ $$ S(\psi^* \phi, \psi^* g) = S(\phi,g). $$ Now, if we further assume that $\psi$ is a conformal map on the manifold $(M,g)$, then $\psi^* g = e^\omega g$ for some $\omega$. Then $$ S(\phi \circ \psi^{-1},g) = S(\phi, \psi^* g) = S(\phi, e^\omega g) = S(\phi,g)\,, $$ i.e. the action describes a conformally symmetric field. I guess this is the reason why Weyl+Diffeo invariance is often the definition of a conformal field theory.
My question is that why do some textbooks and other sources say that "Weyl+Diffeo = Conformal Symmetry in flat background" (for example Erbin's recent String Field Theory book)? Is taking $g$ to be flat necessary?
Why I find this weird is that many Riemann surfaces don't admit a flat metric. (Of course Weyl+Diffeo invariance makes sense in any surface, so maybe that's the reason it is a better definition of conformal symmetry)
I wonder if in my computation above something is wrong with saying that $\psi^*g = e^\omega g$. Certainly this is true for Euclidean metric $g=\delta$ and conformal $\psi$, but for arbitrary $g$ the pullback changes the point where we evaluate $g$ $$ \psi^* g = (D\psi)^t (g\circ \psi) D \psi\,, $$ but it seems that one could still take $e^\omega = |\psi'(z)|^2 (g_{z \bar z}\circ \psi)/ g_{z \bar z}$ in conformal coordinates.
