1. 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.

2. 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}

1. 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.

2. 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}