concerning the first question, related to the German text by Matthias Gaberdiel (greetings to him):
Closed algebra implies symmetry
It is enough to construct the generators of an algebra - in this case, conformal algebra - and calculate their commutators $[L_m,L_n]$ and so on. If the commutators are linear combinations of other generators, we say that the generators form a closed algebra. Now, you are right that we also want to use some "dynamical information about the theory". You wrote that the action should be invariant under the transformations that these generators generate - and you worry that the action - all the dynamical information about the theory - has been completely removed from the proof, right?
That's a good point but the dynamical information hasn't been removed because a particular generator (or, for a general basis of generators, a linear combination of generators) is the Hamiltonian that determines the dynamics itself. For conformal symmetry, it is $L_0+\tilde L_0$ that plays the role of the Hamiltonian. It generates translations of the cylinder or, equivalently (as we will discuss below), multiplicative translations in the radial coordinate. (Maybe additive shifts such as $c/24$ should be added on one of the backgrounds.)
Because $L_0+\tilde L_0$ is a linear combination of some generators and you can show that the set of generators is closed under the operation of taking the commutator, it proves that the whole algebra generated by this set of generators is a dynamical symmetry. In this case, the commutator $[L_m, L_0+\tilde L_0]$ is not strictly zero - so the generators $L_m$ don't commute with the Hamiltonian. Instead, the commutator is equal to another combination of the symmetry generators. But we still say that $L_m$ is a symmetry of the system and we know this situation from other contexts as well.
For example, in special relativity, the angular momentum $J_{12}=J_z$ commutes with the energy $p_0$. However, the Lorentz boost generator $J_{03}$ doesn't commute with the Hamiltonian $p_0$: their commutator is proportional to $p_3=p_z$, a component of the momentum. It's nonzero but it's another symmetry generator. It's normal for symmetry generators that act nontrivially on time - such as the $J_{03}$ boost generator in relativity or $L_m$ in conformal symmetry - to have nonzero commutators with the Hamiltonian $p_0$ or $L_0+\tilde L_0$, respectively. What's important is that the commutator is another operator we know to be a generator of a symmetry, and the symmetry algebra is fully described by group theory - by the structure constants $f$ in $[L_m,L_n]=f_{mn}^k L_k$ - and doesn't need us to know any detailed dynamical information about the fields etc.
You proposed that one should be verifying that the action is invariant under symmetry generators. That sounds good except that the action is only good for a classical description - or a quantum description that is obtained by a direct quantization of a classical theory. Such a way to obtain a quantum theory is only smooth or useful if the quantum theory is "close enough" to a classical theory. The most general CFT, especially in 2 dimensions, is so strongly "quantum" that there is no natural notion of an action and classical degrees of freedom. One must directly work with the quantum operators, their commutators, and they don't have any natural or helpful classical limit. Take the Ising model CFT as an example. You will find lots of fields, such as spin fields and twist fields, whose (mass) dimensions are fractional numbers such as $1/16$, something that would be unthinkable in a classical theory: the whole dimension comes from quantum effects. That's why the structure of the CFT science tries to be as independent of classical concepts such as the action as possible.
Implementing a symmetry on operators
If you have a generator $G$ of a Lie symmetry, it (infinitesimally) acts on ket states $|\psi\rangle$ and bra states $\langle \varphi|$ as
$$\delta |\psi \rangle = i \epsilon G |\psi \rangle, \quad \delta \langle \varphi| = i \epsilon \langle \varphi| G$$
If you also define the action of the generator $G$ on a general operator $M$ as
$$\delta M = i \epsilon [G,M]$$
then you may prove that all matrix elements will be invariant under the symmetry,
$$\delta \langle \varphi | M | \psi \rangle = 0$$
by the Leibniz rule. So the natural action of symmetry generators such as $L_m$ on operators such as $\phi(z,\bar z)$ is
$$\delta \phi(z,\bar z) = i \epsilon [L_m,\phi].$$
So if the commutator of $L_m$ with some fields - operators - is the same as the appropriate $(n+1)$-st derivative of these operators, then the generators $L_m$ implement the symmetry whose infinitesimal form involves $\delta \phi\sim z^{n+1} \partial^{n+1} \phi$.
Your comments about "eigenvalues" are conceptually misguided because a defining property of an "eigenvalue" is that it must be a "value" - a $c$-number - but $\partial_z$ is not a value - it is an operation. (I avoided the word "operator" because $\partial_z$ is not an operator acting on the Hilbert space of the CFT; only operators such as $\phi(z,\bar z)$ and $\partial_z \phi(z,\bar z)$ or $L_m$ are operators acting on the CFT Hilbert space. Instead, $\partial_z$ itself is just a rule to produce one operator from another one. It would be an operator if the wave functions - state vectors - were equivalent to functions of $z,\bar z$ but in a two-dimensional CFT, they surely aren't.)
Why cylinder is important
Concerning the question based on David Tong's text (greetings to David!), the cylinder is important exactly because a CFT on a cylinder is exactly equivalent to a CFT on the infinite plane. If $w=\sigma+i\tau$ lives on a cylinder - with $\sigma$ being $2\pi$-periodic - and if $z=\exp(-iw)$, then the infinite cylinder will be fully mapped to the plane, in a one-to-one way.
I actually think that David is being very clear about it.
So the analysis of the CFT defined on a full plane in general, and its behavior near the $z=0$ origin in particular, is totally equivalent to an analysis of a CFT defined on a cylinder in general, and in the limit $w\to -i\infty$ in particular. The two problems are exactly equivalent, exactly because of the conformal symmetry. The cylinder has a periodic spatial coordinate but this periodicity is not postulated for ad hoc reasons. It's postulated because if you write $z$ in the radius/phase form,
$$ z= \exp(-i\sigma+\tau), $$
then the points with $\sigma\sim \sigma+2 \pi$ are identified with one another. The coordinate $\sigma$ is periodic. This is the basic fact of the exponential function - or its inverse function, the logarithm, if viewed as a function of a complex variable. And the exponential conformal map is very useful which can be seen if you follow what David is doing with that. You could ban the exponential function because you don't like it (or you think that other functions are being discriminated against) - but then you couldn't learn much of the CFT calculus because the CFT calculus largely depends on this clever exponential conformal map.
Because a small piece of any two-dimensional world sheet - regardless of the topology - looks like the flat plane and because the flat plane is equivalent to the infinite cylinder, the infinite cylinder is important for the understanding of local physics of the CFT at any Riemann surface - of any topology.
It is just true that I may just describe the infinite plane in coordinates such that one of them is periodic. This is what makes the analysis of the states defined on the cylinder - closed string states - automatically useful for the analysis of any properties of CFT, including its operators on the plane. In fact, the states of a closed string - obtained by quantizing the CFT on a cylinder - are in one-to-one correspondence with the local operators $\phi_K(0)$ at the origin (or any other point), because of the very same conformal map from the plane to the cylinder.
Now, you ask, where is the quantization?
Many of the formulae would work for a classical (non-quantum) conformal field theory, too. However, there are is no Hilbert space of "states" of a closed string, obtained from the quantization. So many of the interesting things, including the state-operator correspondence discussed two paragraphs above this one, only arise in the quantum theory. Pretty much all the objects such as $H$, $T_{\rm cylinder}$, and so on that David lists on page 86 or almost any other page are operators, so he deals with a quantum theory.
In some equations, David surely also uses commutators, to prove that it is a quantum theory, but it's not necessarily page 86 or another page you could find that has no commutators on it. ;-) But your complaint that David doesn't play with commutators of some fields exactly on some page where you would expect it surely not a sensible complaint, is it?
I am pretty sure that if you listen carefully, you will also understand that the commutators of operators in a CFT may be obtained from the OPEs, the operator product expansions. Just place two operators $T_k(z)$ and $T_l(0)$ to two nearby points $0$ and $z$ and calculate their product. The product will typically include a singularity that diverges as $z\to 0$ - as the two operators are very close to each other. (The singularity will be visible in any sensible expectation value.) The coefficient of $1/z$ or $1/z^2$ or $1/z^4$ - the leading singularity - is either a $c$-number or another operator. From that operator, you may determine the commutator of Fourier modes of $T_k$ and $T_l$ expanded over the cylinder, and so on.
Quantum mechanics has many effects that you wouldn't encounter in classical physics. For example, as you correctly mention, it influences the transformation from the cylinder to the plane, and so on. However, I don't know what to do with questions such as "And I don't see where this is happening here?" What should be happening here? Well, what's happening is probably something else than what you were expecting to be happening - but that's the very reason why you're trying to learn new things from David Tog, isn't it? If you were only learning old things that you knew, you would be wasting your time.
The things that you have to learn to understand two-dimensional conformal field theories are not "the same things" that you already learned for a generic quantum field theory in a generic flat space (such as a four-dimensional one). It is a new topic with new special features such as the exponential maps, OPEs, state-operator correspondence, and so on, and you shouldn't insist that the physics of OPEs has to be composed of the same insight that you already knew from QED in $d=4$. It is not the same thing - if it were the same thing, people wouldn't teach it twice.
So I would suggest that you ask about some particular statements that David makes and that you don't understand. A necessary assumption is that you actually try to listen what David is saying, instead of trying to force him to say things that you wanted to hear in the first place. ;-) When you switch to this mode of learning, the discussion could become a little bit more constructive. At any rate, I assure you that David is talking mostly about quantum mechanical systems, so all observables are operators on a Hilbert space that can get multiplied and whose expectation values may be calculated. The previous sentence could help if you misunderstood every single formula in David's lectures that includes an operator - which would be pretty much every formula.
However, I can't explain you all other details about David's text (and not even all effects of quantum mechanics - because pretty much everything in the text is quantum mechanical) unless you say exactly what's your problem. I would have to take 107 of his pages, inflate them by a factor of 10, and you could still end up being dissatisfied because your dissatisfaction could have some totally different causes. ;-)
