I think it is just not very well written.
For the holomorphic part $T(z)$, we should have weights $ (h,\bar h) =(2,0)$, and for the anti-holomorphic part $\tilde T(\bar z)$, we should have weights $ (h,\bar h) =(0,2)$, so for each part, we should have a scaling dimension $2$, and a spin $\pm 2$
However, the holomorphic and anti -holomorphic parts of the Energy-momentum tensor are not generally primary fields (see $5.124$ p $136$, $5.121$ p $135$) because of a possible central charge.
The significance of the Möbius transformations $\mathrm{PSL}(2,\mathbb{C})$ in 2D conformal field theory is that they are the globally defined conformal transformations on the Riemann sphere.
While the infinitesimal conformal transformations form the infinite-dimensional Witt algebra spanned by the vector fields
$$ L_n = -z^{n+1}\partial_z$$
we must be mindful that those vector fields are not globally defined on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$. Obviously, they are singular at $z = 0$ for $n < -1$. Changing coordinates by $z\mapsto w = \frac{1}{z}$, we get
$$ L_n = -w^{1-n}\partial_w$$
which is singular at $w=0$, i.e $z=\infty$, for $n > 1$.
Therefore, the only globally defined conformal generators are $L_{-1},L_0,L_1$. These three generate precisely the group of Möbius transformations $z\mapsto \frac{az+b}{cz+d}$.
Thus, the symmetry group of a conformal field theory on the Riemann sphere is just $\mathrm{PSL}(2,\mathbb{C})$, and we have the requirement that the stress-energy tensor also should be invariant under this symmetry group. No such requirement can be said for the infinitesimal transformation of the Witt algebra. Nevertheless, classically, the stress-energy tensor transforms with its usual conformal weight also under those, since there is no central charge.
In the course of quantization, we incur a central charge for the Witt algebra, turning it into the Virasoro algebra1. Since the energy-momentum tensor is $T(z) = \sum_n L_n z^{n-2}$, the appearance of the central charge means the classical transformation law under the infinitesimal transformations generated by the $L_n$ may change by a quantum correction - this is precisely the Schwarzian derivative term. In the classical case $c = 0$, it vanishes, as a quantum correction (in this case often interpreted as a normal ordering constant) should.
However, if this also changed its behaviour under the global transformations, then the quantum theory would become anomalous, in particular, it would break the conservation of the Noether currents associated to $L_{-1},L_0,L_1$, which are $T(z),zT(z),z^2T(z)$. That is, anomalous transformation under a $\mathrm{PSL}(2,\mathbb{C})$ transformation would break energy-momentum conservation. This is undesirable, and, in fact, does not happen (as you may convince yourself by just chugging through the calculation of the transformation behaviour of $T$).
Now, why does the Schwarzian derivative appear as the quantum correction? If you start from the requirement that the quantum correction must vanish for $c=0$ and for $\mathrm{PSL}(2,\mathbb{C})$ transformations, then it is clear that it must be proportional to $c$. Furthermore, whatever $\{z,w\}$ is, it has to respect the group composition law that two successive transformations $z\mapsto w \mapsto u$ give the same as mapping $z\mapsto u$ directly. This is equivalent to the equation
$$ \{u,z\} = \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}z}\right)^2\{u,w\} \tag{1}$$
since
$$ T(u) = \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(T(w) + \{u,w\}\right)$$
and
$$ T(w) = \left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right)$$
but also
$$ T(u) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$
so we obtain
$$ \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(\left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right) + \{u,w\}\right) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$
which gives
$$ \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2\{u,w\} = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{u,z\} $$
after subtracting $\left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 T(z)$ from both sides. Multiplying by $\left(\frac{\mathrm{d}u}{\mathrm{d}z}\right)^2$ now yields eq. (1).
It can be shown that $(1)$ together with the requirement of $\mathrm{PSL}(2,\mathbb{C})$ invariance define the Schwarzian derivative uniquely.
1Shameless self-promotion: See this Q&A of mine for why we get a central charge in the quantum theory.
Best Answer
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.
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.
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.
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 $T_{zz}(z, \bar z) = T(z)$