Please note that the light-cone coordinate transformation is not an element of the <PERSON> group, nor of the full Conformal group. The easiest way to see this for the conformal group is to consider the transformation properties of the metric tensor. For the <PERSON> group note that the proper distance $x_0^2-x_1^2 \rightarrow t\bar{t}$ which is not of the form $x_0^2-x_1^2 \rightarrow {x'}_0^2-{x'}_1^2$.
Therefore the scalar field $\Phi_a(x)$does not transform in a representation of some corresponding group when moving from $(x_0,x_1)$ to $(t,\bar{t})$.
The way I would go about deriving eqn. $(1.2.5c)$ and $(1.2.5d)$ (which are the correct forms of the respective commutators) would be in exact analogy with the steps required to go from eqn. $(1.1.5)$ to eqn. $(1.1.8)$. 364
Namely, one starts with: \begin{equation} U(\gamma)\Phi_a(t,\bar{t})U(\gamma)^{-1} = (ct+d)^{-2\Delta_a}(\bar{c}\bar{t}+\bar{d})^{-2\bar{\Delta}a}\Phi(\gamma(t,\bar{t})) \tag{1.2.4} \end{equation} Note that exactly as in equation $(1.1.8)$ where $-Q_k$ is the operator corresponding to the special conformal transformation parametrised by the parameter $q_k$, in this case the operator $-Q(-\bar{Q})$ corresponds to special conformal transformation parametrised by $b+(b_-)$ (ref. eqn. (1.2.3)).
Therefore the Lie Group elements $U(\gamma)$ are given by:
$$ U(\gamma)= e^{-i(b_+Q+b_-\bar{Q})} $$
In the following I drop the $\bar{t}$ dependence here since the treatment is exactly analogous, and the left and right moving (holomorphic and anti-holomorphic) components decouple. Now, eqn. $(1.2.4)$ gives:
$$ e^{-ib_+Q}\Phi_a(t)e^{ib_+Q} = (1+b_+t)^{-2\Delta_a}\Phi_a(\frac{t}{1+b_+t}) $$
expanding as a series in $b_+$ $$ (1-ib_+Q+\dots)\Phi_a(t)(1+ib_+Q+\dots) = (1-2\Delta_ab_+t+\dots)(\Phi_a(t)-b_+t^2\partial_t\Phi_a(t)+\dots) $$
Collating terms in powers of $b_+$: $$ \Phi_a(t)-ib_+[Q,\Phi_a(t)]+\mathcal{O}(b_+^2) = \Phi_a(t)-b_+(t^2\partial_t+2\Delta_at)\Phi_a(t)+\mathcal{O}(b_+^2) $$
and comparing the coefficients of $b_+$ one gets: $$ i[Q,\Phi_a(t)] = (t^2\partial_t+2\Delta_at)\Phi_a(t) $$
Finally, remember that in 2 dimensions, there are in general 2 distinct scaling dimensions ($\Delta_a$ and $\bar{\Delta}_a$) which are not the same as the scaling dimension $\Delta_a$(Let's call it $\Delta_a'$ for clarity) in eqn. $(1.1.5)$. These satisfy $\Delta_a+\bar{\Delta}_a=\Delta'_a$ and only in case of a spinless field have $\Delta_a$ = $\bar{\Delta}_a$. 804