I thought at first this question was more general involving the nature of the Virasoro algebra. As a result the first two paragraphs are boiler plate discussions on that. The actual question is addressed in the third paragraph.
The vector fields $T^m~=~z^{m+1}\partial z$ satisfy the Witt algebra or Virasoro algebra without central extension for each index value,
$$
[T^m,~T^n]~=~(n~-~m) T^{m+n}.
$$
The Virasoro algebra maybe extended by a one-dimensional center with the map
$$
{m:Vir\oplus {\mathbb C}~\rightarrow~Vir,}
$$
where this central extension is the kernel of the map. The extended Virasoro group is then,
$$
[T^m,~T^n]~=~(n~-~m)T^{m+n}~+~c(m,~n).
$$
This is imposed since with the case $m~+~n~=~0$ then $[T^m,~T^{-m}]~=~-2mT^{2m}$ and there are two infinite sums with the same result, one for $m~\ge~0$ and $m~\le~0$. To give a central extension for a specific index $m$ this is a c-number in the $\mathbb C$ in the domain of the above map, and $c(m,~n)~=~c(m)\delta_{m+n}$ The central extension may then be computed with the Jacobi identity
$$
[T^k,~[T^m,~T^n]]~+~ [T^m,~[T^n,~T^k]]~+~ [T^n,~[T^k,~T^m]]~=~0.
$$
which is equal to
$$
(m~-~n)c(k)~+~(n~-~k)c(m)~+~(k~-~m)c(n)~=~0.
$$
To get $c(m)$ we set $k~=~1$ and $n~=~-(m~=~1)$. This gives us
$$
c(m~+~1)~=~\frac{(m~+~2)c(m)~-~(2m~+~1)c(1)}{m~-~1}.
$$
This is a nice recursion relationship and we get $c(2)~=~3c(1)~-~3c(1)$ $=~0$ and the general recursion
$$
c(m)~=~c(3)m^3~-~c(1)m
$$
The $T^m$ are expanded according to string modes
$$
T^m~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{m-n}\alpha_n.
$$
Form here the coefficients $c(3)$ and $c(1)$ can be determined. The $T^0,~T^{\pm 1}$ generate a closed $SL(2,{\mathbb R}$ subalgebra. The $T^0~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{-n}\alpha_n$ annihilates the vacuum state as does $L^{\pm}$. For $m~=~2$ one can show by computing the vacuum expectation $\langle 0|T^{-2}T^2|0\rangle$
$$
c(m)~=~\frac{D}{12}(m^3~-~m)
$$
with some additional work, which is a bit lengthy, you can show $D~=~26$ and work up the bosonic string.
Any field $\phi$ transforms under the conformal algebra as its commutator with a $T^m$ element of is $iz^{m+1}\partial_z\phi$. The $SL(2,{\mathbb R})$ is the symmetries of conformal quantum mechanics, and the Witt algebra or Virasoro algebra is then its generalization as the set of diffeomorphisms of the circle. Suppose a function or field that transforms under these diffeomorphisms requires this central extension. For instance the Hamiltonian is $H~=~T^0$ and suppose the field is $\phi(\tau)~=~\phi(0)e^{iH\tau}$, then
$$
\delta\phi~=~\delta\phi(0)e^{iT^0\tau}~+~i\phi(0)\epsilon\sum_m[T^m,~T^0]~+~O(\tau^2).
$$
It is then clear that the central extension is $c(0)~=~0$. It is not hard to see that a similar situation hold if the field is dependent on $T^{\pm}$. However, if the field depends upon $L^m$, for $|m|~>~1$ then care needs to be taken. The $L_m$ are the Fourier modes of the stress-energy tensor in the string $1~+~1$ dimensional spacetime. So it is unclear whether a field is of this nature. I would be tempted at this point to say no, except for the case of the $T^0$, which is the Hamiltonian. However, in that case the central or anomaly term is zero. The only reason one would need to be concerned with the central term is if $\phi$ is an element of the circle diffeomorphism, or is some function of it.
