In the case of a matrix Lie group and a projective representation, central charge is defined here as follows.
If $M$ is a matrix Lie group, then elements $G$ of its Lie algebra $\mathbf{m}$ can be given by
$$G = \frac{d}{dt}\left .(g(t))\right|_{t=0},$$
where $g$ is a differentiable path in $M$ that goes through the identity element at $t=0$. Commutators of elements of the Lie algebra can be computed using two paths, $g_1$ and $g_2$, and the group commutator, $$ [G_1, G_2] = \frac{d}{dt} \left .g_1(t)g_2(t)g_1(t)^{-1}g_2(t)^{-1}\right|_{t = 0}, \quad G_1 = g_1'(0), G_2 = g_2'(0).$$
Likewise, given a group representation $U(M)$, its Lie algebra $\mathbf u(\mathbf{m})$ is computed by
$$\begin{align} [U_1, U_2] &= \frac{d}{dt} \left .U(g_1(t))U(g_2(t))U(g_1(t))^{-1}U(g_2(t))^{-1}\right|_{t = 0}\\ &= \frac{d}{dt} \left .U(g_1(t)g_2(t)g_1(t)^{-1}g_2(t)^{-1})\right|_{t = 0}, \quad G_1 = g_1'(0), G_2 = g_2'(0).\end{align}$$
Then there is a Lie algebra isomorphism between $\mathbf{m}$ and $\mathbf u(\mathbf{m})$ sending bases to bases, so that $u$ is a faithful representation of $\mathbf{m}$.
If, however, $U(G)$ is a projective representation, i.e., a representation up to a phase factor, then the Lie algebra, as computed from the group representation, is not isomorphic to $\mathbf{m}$. In a projective representation, the multiplication rule reads $$U(g_1)U(g_2) = \omega(g_1, g_2)U(g_1g_2) = e^{i\xi(g_1, g_2)}U(g_1g_2).$$ The function $\omega$, often required to be smooth, satisfies $$\begin{align}\omega(g,e)&=\omega(e,g) = 1,\\ \omega(g_1, g_2g_3)\omega(g_2,g_3) &= \omega(g_1,g_2)\omega(g_1g_2,g_3)\\ \omega(g,g^{-1})&=\omega(g^{-1},g).\end{align}$$ It is called a 2-cocycle on $M$.
One has $$\begin{align} [U_1, U_2] &= \frac{d}{dt} \left .U(g_1(t))U(g_2(t))U(g_1(t))^{-1}U(g_2(t))^{-1}\right|_{t = 0}\\ &= \frac{d}{dt} \left .e^{i\xi(g_1,g_2)\xi(g_1^{-1}, g_2^{-1})\xi(g_1g_2, g_1^{-1}g_2^{-1})}U(g_1(t)g_2(t)g_1(t)^{-1}g_2(t)^{-1})\right|_{t = 0}\\ &\equiv \frac{d}{dt} \left .\Omega(g_1,g_2)U(g_1(t)g_2(t)g_1(t)^{-1}g_2(t)^{-1})\right|_{t = 0}\\ &= \left .\frac{dU(g_1(t)g_2(t)g_1(t)^{-1}g_2(t)^{-1})}{dt}\right|_{t = 0} + \left .\frac{d\Omega(g_1,g_2)}{dt}\right|_{t=0}I, \quad G_1 = g_1'(0), G_2 = g_2'(0),\end{align}$$ because both $\Omega$ and $U$ evaluate to the identity at $t=0$. For an explanation of the phase factors $\xi$, see Wigner's theorem.
The commutation relations in $\mathbf{m}$ for a basis, $$[G_i,G_j] = {C_{ij}^k}G_k,$$ become in $u$, $$[U_i,U_j] = {C_{ij}^k}U_k + D_{ij}I,$$ so in order for $u$ to be closed under the bracket (and hence have a chance of actually being a Lie algebra), a central charge $I$ must be included.
I note that they probably wanted to write in the last row
a central charge $D_{ij}I$
instead of
a central charge $I$
But how does it go when $G$ is a general Lie group instead of being a matrix group? One problem is that in the case of a general Lie group - in contrast to a matrix group - we cannot apply the Leibniz rule for derivating a product. Also confusing to me that according to other resources, the central charge should appear in the commutator in the central extension of $G$ while here it appears in $[U_j,U_j]$. Now, which one is the real central charge and what is the other?