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.
I'm going to try to get at the crux of your questions without worrying too much about mathematical rigor/details (as is the physicist's way), but hopefully there are enough details so that the answer is clear.
Why does this have anything to do with energy or momentum?
First, a bit of background. In physics, a theory of fields $\phi$ on a manifold $M$ is often specified by an action $S$; a functional which maps a given field configuration $\phi$ to a number (often the target set of the action is either $\mathbb R$ or $\mathbb C$). For, concreteness, consider a field theory on $\mathbb R^d$. As it often turns out, the action of such a field theory is translation invariant. This means that if we define the action of the group of translations of $\mathbb R^d$ on the fields $\phi$ of the theory by $\phi\to\phi_\epsilon$ where
$$
\phi_\epsilon(x) = \phi(x-\epsilon)
$$
then
$$
S[\phi] = S[\phi_\epsilon]
$$
In such cases, a theorem in field theory called Noether's theorem guarantees the existence of a conserved tensor $T^{\mu\nu}$ associated with this invariance, namely one for which
$$
\partial_\mu T^{\mu\nu} = 0
$$
This conserved tensor associated with translation invariance of the action is what we call the energy-momentum tensor, and this is essentially the tensor we're talking about in the context of conformal field theory.
So what the heck does this object having anything to do with energy and/or momentum? Well, we can motivate this physically through examples. If you take, as an example of a field theory, electromagnetism, then you find that the components $T^{\mu\nu}$ of the energy-momentum tensor physically represent quantities like the energy density stored in the fields. One finds, for example, that the $00$ component of the electromagnetic energy-momentum tensor has the expression
$$
T^{00} = \frac{1}{8\pi}(\mathbf E^2 + \mathbf B^2)
$$
which one can show, by other means, is precisely the physical energy density stored in the electromagnetic fields.
Conformal invariance is implied by the condition $\mathrm{tr} \,T=a_1+a_3=0$. (What does this sentence mean?)
One can show that under a coordinate transformation $x\to x+\epsilon(x)$, the action of a sufficiently generic field theory transforms as
$$
S \to S+\frac{1}{2}\int d^dx \,T^{\mu\nu}(\partial_\mu\epsilon_\nu + \partial_\nu \epsilon_\mu) + O(\epsilon^2)
$$
A conformal transformation has the property that
$$
\partial_\mu\epsilon_\nu + \partial_\nu \epsilon_\mu = \frac{2}{d}\partial_\rho \epsilon^\rho \,\delta_{\mu\nu}
$$
which gives
$$
S \to S+ \frac{1}{d}\int d^dx\, T^\mu_{\phantom\mu\mu}\partial_\rho\epsilon^\rho + O(\epsilon^2)
$$
Notice that the integrand contains the trace $T^\mu_{\phantom\mu\mu}$ of the energy-momentum tensor, and we see that if this trace vanishes, then the action has the property
$$
S \to S+ O(\epsilon^2)
$$
It is invariant to first order in $\epsilon$. This is a sort of "infinitesimal invariance" as a physicist might call it, and it is what the statement is referring to in this context.
What type of object is $T(z)$? And how is obtained from $T$?.
For a conformal field theory on $\mathbb R^2$, after going to complex coordinates $z,\bar z$ it is possible to show that $\partial_{\bar z} T_{zz}(z,\bar z) = 0$, so for the sake of notational compactness, one often writes $T_{zz}(z, \bar z) = T(z)$
Best Answer
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 $T = 0$.