CFT consists for most mathematicians - who are interested in this topic - currently of the study of vertex operator algebras, see this question on math overflow:
You can find a little bit more about the topic and the work of several mathematicians here:
As you can see from the answers on mathoverflow, vertex algebras were not invented for the study of CFT, and that they form an axiomatic abstraction of operator algebra products was noted only later.
A personal and very subjective note: One should not underestimate the amount of theoretical physics that is necessary to understand what a QFT is to physicists. Most mathematicians that encounter physics for the first time since highschool through some QFT framework seem to be quite taken aback by the high intrance fee they'd have to pay to understand this. This is my own, personal explanation for the observtion that most mathematicians study the formal machinery only, in order to use it to prove some new mathematical theorems, but only very rarely in order to better understand what physicists do. Although you'll find quite a lot of work by quite a lot of rather famous mathematicians if you follow the links above, AFAIK there is none doing the work you describe.
First, the full paper is here:
http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=807BE383780883ACB4CAB8BD48E8C90B?doi=10.1.1.128.1806&rep=rep1&type=pdf
Second, the paper has 150 citations. See all this information at INSPIRE (updated SPIRES):
http://inspirebeta.net/record/278923?ln=en
Third, the text between 3.4 and 3.5 looks totally comprehensible. At that point, they are able to define $n\cdot S$ modulo 1, which is equivalent to defining the action $S$ modulo $1/n$. The goal is to define the action $S$ itself modulo 1; I suppose that their normalization of the path integral has to have $\exp(2\pi i S)$ with the atypical $2\pi$ factor. Yes, confirmed, it's equation 1.2.
If you shift the action by 1 - or $2\pi$ in the ordinary conventions - it doesn't change the integrand of the path integral; it doesn't change the physics. So quite generally, if one is able to say that the action $S$ is equal to $S_0+n$ (or $2\pi n$ normally) for some integer $n$, he knows everything about the physics of the action he needs; shifting it by an integer doesn't change anything. That's why, in fact, the action is often defined modulo 1 only (up to the addition of an integer multiple of 1).
So it's enough to know the "fractional part" of the action; the integer part is irrelevant. However, at the point of the equation 3.4, their uncertainty is larger than that: they only know the action modulo $1/n$. For example, if the action is $9.37$ modulo $1/2$, it means that the fractional part may be $0.37$ but it may also be $0.87$. These two values of $S$ would change the physics because the contribution of the configuration to the path integral changes the sign if one changes $S$ by $1/2$ (in normal conventions, by $\pi$).
If one only knows $S$ modulo $1/n$, and if he thinks it's $S_0$ - in this case, the $F\wedge F$ expression - it means that the real action is
$$ S = S_0 + K/n $$
and the integer $K$ has to be determined. Because the change of the action $S$ by an integer doesn't change physics, it doesn't matter if $K$ in the equation above is changed by a multiple of $n$. So the goal is to find the right $K$ to define the action - and $K$ is an unknown integer defined (or relevant) modulo $n$, i.e. up to the addition of an irrelevant and arbitrary multiple of $n$.
At some point, they find the right answer and it is
$$ K = -\langle \gamma^*(\omega),B\rangle $$
which removes the ambiguity of $S$ - the missing knowledge whether $S$ should be the original $S$ or higher or smaller by a particular multiple of $1/n$. If you don't understand the text above, then apologies, I have no way to find out why, so I can't give you a better answer unless you improve your question.
Best Answer
The work being done on TQFT by physicists and mathematicians is wonderful, but in no way should you think it somehow captures what is important in QFT to physicists trying to explain the real world. QFT as applied to the real world has particle-like excitations, non-trivial correlation functions depending on the spacetime distance between operators, spontaneous symmetry breaking, non-trivial renormalization group flows and so on and calculations are done in 4d flat Minkoswki space because this is an excellent approximation to the local spacetime metric. TQFT, as the name implies, captures information about the topology of manifolds and would be completely boring on $R^{3,1}$. You can't use it to study pion scattering, or compute the short distance interaction between quarks, or the production cross-section for the Higgs boson or really anything that particle theorists do with QFT. So the answer to your first question is no. The axioms of TQFT do not capture what physicists feel is important in QFT. It isn't really a question of adding more parameters, they are just very different beasts. To your question of whether mathematicians have gotten rid of too much for physicists to care, the answer is yes, for most physicists. As I said earlier, for all practical purposes on can do particle theory on flat Minkowski space and the only structure that really matters is the causal structure. None of this is meant to denigrate work on TQFT of course.