You don't need to use that. You can simply do the cross-contractions by hand. Let's do that. Note that I only care about the the $\frac{1}{z^4}$ term to evaluate the central charge. We have
\begin{equation}
\begin{split}
T(z) T(w) & =\left( : \partial_z b c(z): - \lambda \partial_z : b c (z): \right) \left( : \partial_w b c(w): - \lambda \partial_w : b c (w): \right) \\
& = : (\partial_z b) c(z): : (\partial_w b) c(w): - \lambda \partial_z : b c (z): : (\partial_w b) c(w): \\
&~~~~~~~~~~~~~~~~~~~~~~- \lambda : (\partial_z b) c(z):\partial_w : b c (w): + \lambda^2 \partial_z : b c (z):\partial_w : b c (w):
\end{split}
\end{equation}
Now at each step, we only keep the full contractions to extract the central charge. We then find
\begin{equation}
\begin{split}
T(z) T(w) &\sim \partial_z \frac{1}{z-w}\partial_w \frac{1}{z-w} - \lambda \partial_z \left( \frac{1}{z-w} \partial_w \frac{1}{z-w} \right)\\
&~~~~~ - \lambda \partial_w \left( \frac{1}{z-w} \partial_z \frac{1}{z-w} \right) + \lambda^2 \partial_z \partial_w \frac{1}{(z-w)^2} \\
&= \frac{-6\lambda^2 + 6 \lambda - 1 }{(z-w)^4} + \cdots
\end{split}
\end{equation}
We can then read off
$$
c = 2 \left( -6\lambda^2 + 6 \lambda - 1 \right) = - 3 (2 \lambda - 1 )^2 + 1
$$
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
The central charge itself appears in the commutation relations because the quantum theory allows not only linear, but projective representations of the conformal Witt algebra to be physically admissible. Such projective representations are in bijection to linear representations of the central extensions of the Witt algebra $$[L_m,L_n] = (m-n)L_{m+n}$$ which are the Virasoro algebras $$ [L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}(m^3-m)\delta_{m,-n} $$ for $c\in\mathbb{R}$. For more on projective representations and central extensions, see this Q&A of mine.
The energy-momentum tensor classically has conformal weight $2$, so its conformal mode expansion is $T(z) = \sum_n T_n z^{-n+2}$. The question now becomes how one shows that $T_n = L_n$. Since the energy-momentum tensor is the conserved current for the translations generated by $L_{-1} = \partial_z$ (since $L_n = z^{-n+1}\partial_z$), its integral has to be the generator $L_{-1}$ itself (the integral of the conserved current is the Noether charge, which is the generator of the symmetry in the Hamiltonian formulations): $$ L_{-1} = \frac{1}{2\pi\mathrm{i}}\int T(z)\mathrm{d}z$$ and inserting the mode expansion we get $L_{-1} = T_{-1}$. More generally, for any conformal transformation $z\mapsto z+\epsilon(z)$, we get a conserved current $\epsilon(z)T(z)$. The $L_n$ generate the transformations $z\mapsto z + \epsilon_n z^{n+1}$. As above, it follows that $$ L_n = \frac{1}{2\pi\mathrm{i}}\int z^{n+1}T(z)\mathrm{d}z$$ and thus $L_n = T_n$. Therefore, the energy-momentum tensor of a conformal field theory is $T(z) = \sum_z L_n z^{-n+2}$.
Here we only considered the holomorphic parts, but the arguments for the anti-holomorphic pieces are exacctly the same