The conformal group is defined for any spacetime you want. The conformal group of d-dimensional Euclidean space, which has isometry group SO(d), is SO(d+1,1). The conformal group of d+1 dimensional Minkowski space, whose isometry group is SO(d,1), is SO(d+1,2). The defining property of the conformal group is that its the set of transformations that leave the metrix $g_{\mu\nu}$ invariant up to a scaling factor $e^{\omega(x)}$.
A more general definition for CFTs in d>2 dimensions is that a CFT is a QFT whose Hilbert space breaks up into representations of the conformal group and whose correlation functions are invariant under any conformal transformation (I don't believe we have to restrict to primary operators).
The special case of two dimensions is that the conformal algebra is infinite dimensional. The group of globally defined conformal transformations is still finite, but there are an infinite number of local conformal transformations. So 2d CFTs are a lot more restricted then higher dimensional CFTs.
For a good introduction look at
https://sites.google.com/site/slavarychkov/
Note that under an infinitesimal change in the metric of the form $g \to g + \delta g$ the action changes to
$$
\delta S = \int T^{ab} \delta g_{ab}
$$
Now, under Weyl transformations we have
$$
g_{ab} \to e^{2\omega} g_{ab} \qquad \implies \qquad \delta g_{ab} = 2 \omega g_{ab}
$$
For Weyl transformations $\omega$ is completely arbitrary. If we consider a conformal transformation then, the metric also transforms as above except that $\omega = \frac{1}{d} \nabla_a \xi^a$ where $\xi^a$ is a conformal killing vector, i.e. $\omega$ takes a specific functional form.
Either way, for both conformal or Weyl transformations $\delta g_{ab} =2\omega g_{ab}$. Thus, for either of these transformations, the variation in the metric is
$$
\delta S = 2 \int \omega T
$$
Thus, if the trace of energy momentum tensor vanishes, $T = 0$, then
$$
\delta S = 0
$$
and we have a symmetry of our theory!
OK. So we have shown that if $T = 0$, then the theory is invariant under Weyl and conformal transformations. What about the inverse statement? Can we infer from Weyl and conformal invariance that $T = 0$? The latter is a more subtle question.
Weyl or conformal invariance implies
$$
\int \omega T = 0
$$
Now, when talking about Weyl invariance, the above is true for arbitrary $\omega$. In this case, we can most certainly conclude that $T = 0$ (for instance take $\omega \propto \delta^4(x)$ or some smoothed out version thereof and we immediately reach this conclusion.
When talking about conformal invariance, $\omega$ is not arbitrary and we cannot conclude that $T$ must vanish. For instance, in a flat background, $\omega$ takes the form $\lambda + a_\mu x^\mu$ where $\lambda$ and $a_\mu$ are arbitrary constants. Thus, all we can conclude is that we must have
$$
\int T = 0 ~, \qquad \int x^\mu T = 0
$$
These two conditions no longer imply that $T = 0$. Thus, as per this argument the inverse statement is not necessarily true in conformal field theories. I'm not sure if there is any other argument that can be used to justify that $T$ must vanish in CFTs, but so far, all the CFTs we study have $T = 0$.
Best Answer
It's somewhat unclear how you "understood" and are "happy" about the definition of the conformal transformations because your questions, while referring to page 2 and other things, are nothing else than misunderstandings about the definition of a conformal transformation which is explained on page 1, not 2.
John wrote his equation 1 which states that conformal transformations "are" diffeomorphisms that only change the metric by a local scalar coefficient - by a Weyl rescaling. However, the invariance of a theory under these diffeomorphisms is a trivial property. If one is allowed to change the terms coupled to the "metric", a diffeomorphism-transformed theory has a different action in general, and it is always possible to rewrite the original action in the conformally transformed coordinates.
But what's physically nontrivial is the condition that the action, in its original form, is actually invariant under the operations - that's what we mean by the theory's being conformally invariant. If a theory is conformally invariant, we don't allow any "change of the coefficients" in the integral of the Lagrangian density. This condition of conformal invariance, as he shows, is equivalent to the invariance of the theory under the "Weyl rescaling" only: we just completely eliminate the diffeomorphisms from the picture.
So:
Again, the invariance under the combined "diffeomorphism" and "Weyl rescaling" (the latter changes the form of the action) is a tautology. Obviously, by conformal invariance, we don't mean a tautology, so by conformal invariance, we mean the invariance of the action under the diffeomorphism separately, without changing the form of the metric in the action. Because the invariance under the "combo" is tautological, conformal invariance is equivalent to the invariance under the "Weyl rescaling part" of the transformation only.
No, on page 2, the transformations are exactly what the equations say: point-dependent transformations of the metric tensor itself i.e. a Weyl rescaling. There is no diffeomorphism at this stage. The invariance under those Weyl transformations is equivalent to the invariance under some diffeomorphisms (with the modification of the metric erased), as explained in the previous point.
The formalism may depend on classical physics but your comment that it is "classically only of course" is incorrect, too. All those facts about conformal transformations are completely valid quantum mechanically as well - and indeed, this background is primarily mean to understand some quantum theories.