The answer is quite simple. You should use eqs.(6.8) in the paper. You put them into the last term of eq.(7.11) then, a straightforward integration, I mean something like $\delta\tau\rightarrow\tau$, $\delta A_{\bar w}\rightarrow A_{\bar w}$ and so on, should do the job.

So, let us consider (note that in your post there is a wrong sign)

$$
\delta S=(2\pi)^2 i \left(-T_{ww}\delta \tau+T_{\bar w\bar w}\delta \bar\tau
+\frac{\tau_2}{\pi} J_{w}^I\delta A_{I\bar w}+\frac{\tau_2}{\pi} \tilde J_{\bar w}^I\delta \tilde A_{Iw}\right)_{constant}~.
$$

(here "constant" means that only the zero mode is retained) and the corresponding eqs.(6.8) in Kraus' review

$$\eqalign{ T_{ww}&=-{k \over 8\pi} +{1 \over 8\pi}
A_w^2+{1 \over 8\pi} {\tilde A}_w^2~, \cr T_{{\bar w}{\bar w}}&= -{{\tilde k} \over 8\pi}
+{1 \over 8\pi}A_{{\bar w}}^2+{1 \over 8\pi}{\tilde A}_{{\bar w}}^2~, \cr J^I_w &=
{i\over 2} k^{IJ} A_{Jw}~, \cr {\tilde J}^I_{{\bar w}}& = {i\over 2} {\tilde k}^{IJ}
{\tilde A}_{J{\bar w}}~.}$$

By substitution one has

$$ \delta S = (2\pi)^2 i\left[-\left(-{k \over 8\pi} +{1 \over 8\pi}
A_w^2+{1 \over 8\pi} {\tilde A}_w^2~\right)\delta\tau+\left(-{{\tilde k} \over 8\pi}
+{1 \over 8\pi}A_{{\bar w}}^2+{1 \over 8\pi}{\tilde A}_{{\bar w}}^2~\right)\delta{\bar\tau}\right.$$
$$\left.+\frac{i\tau_2}{2\pi}k^{IJ} A_{Jw}\delta A_{I\bar w}+\frac{i\tau_2}{2\pi}{\tilde k}^{IJ}{\tilde A}_{J{\bar w}}\delta \tilde A_{Iw}\right]_{constant}.$$

From this you get immediately the result when you note that the variation with respect to the gauge field just cancels out the $\tau_1$ contribution, having ${\bar\tau}-\tau$ that comes from the squared terms, and is recovered upon integration.

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