Conformal transformation vs diffeomorphisms

OK I think I know what is going on. It's all about primes. Consider an active spacetime transformation:

$$ x^{\mu} \mapsto x'^{\mu}(x) \, ,$$

$$g_{\mu \nu} (x) \mapsto g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} \, .$$

(the transformation of the metric tensor follows from the fact that it is a rank 2 tensor). With this notation both Di Francesco and David Tong are wrong (as far as I understand). The GR book by Zee on the other hand writes it properly. First of all consider an isometry. This is an spacetime transformation as before that leaves the metric invariant, meaning

$$ g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} = g_{\mu \nu} (x') \, .$$

(watch the primes). On the other hand a conformal transformation is a transformation that satisfies a weaker condition: it leaves the metric invariant up to scale, meaning

$$ g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} = \Omega^2(x')g_{\mu \nu} (x') \, .$$

Now there should be no inconsistency. Di Francesco's definition was wrong (according to this convention/notation/understanding) because it compared the metric before and after the transformation at different points, and you have to compare them at the same point.

I'm a mathematician, not a physicist, so I learned all of these ideas with different notation, but I think I understand what might be confusing you.

Conformal transformations are indeed a special kind of diffeomorphism, and a rotation (say in the plane with the usual metric) is indeed conformal, so the two formulas you listed had better agree in this case.

But in fact, if your manifold is $\mathbb{R}^2$, your metric is the usual one ($g_{\mu\nu}$ is the identity matrix at every $x$), and your coordinate change is a rotation, the second formula you listed will show you that the metric looks unchanged in the new coordinates. (This is not a coincidence: preserving this metric is exactly the property that makes rotations special in the first place!) That is, there is no conflict between the two formulas here, it's just that seeing it involves a bit of computation.

Working this out is a very good exercise and I don't think you'd gain much from me typing it all out here. A hint that might help you get oriented is that, since rotations are linear in the coordinate system we've chosen, the Jacobian matrix at every point is the same as the matrix for the rotation itself.