Willie Wong gives the obvious computational method of computing the desired formula. If you happen to believe the general formula for the derivative of the scalar curvature $R$ given in the first line of Proposition 9, you can save yourself the trouble of going all the way back to the definition of curvature. (NB: I strongly suggest that you derive this formula for yourself if you have never done it, the computation is essentially sketched in Willie Wong's answer. In particular, my answer is not "shorter/easier" unless you already have done the general computation. However, the general formula is quite well known, so perhaps you're comfortable with it already).
The formula for the derivative of scalar curvature of $g$ in the direction of a symmetric $2$-tensor $h$ reads $$ DR|_g(h) = \sum_{i,j=1}^n ((D_g)_{e_i,e_j}^2h)(e_i,e_j) - \Delta_g(tr_g h) - (n-1)tr_g h $$ From this, we see that for the $g(t)$ considered in Proposition 13 $$ g(t) = g_0(t) + t^2\left(\frac{1}{2(n-1)} u \overline g\right) := g_0(t) + t^2h, $$ if we consider the $t^2h$ term as a perturbation of $g_0(t)$ of the form $sh$, then \begin{align*} R_{g(t)} & = R_{g_0(t)} + t^2 DR|_{g_0(t)}h +O(t^4)\\ & = R_{g_0(t)} + t^2\left( \sum_{i,j=1}^n ((D_{g_0(t)})_{e_i,e_j}^2h)(e_i,e_j) - \Delta_{g_0(t)}(tr_{g_0(t)} h) - (n-1)tr_{g_0(t)} h \right)+O(t^4)\\ & = R_{g_0(t)} + t^2\left( \sum_{i,j=1}^n ((D_{\overline g})_{e_i,e_j}^2h)(e_i,e_j) - \Delta_{\overline g}(tr_{\overline g} h) - (n-1)tr_{\overline g} h + O(t)\right)+O(t^4)\\ & = R_{g_0(t)} + t^2\left( \frac{1}{2(n-1)}\Delta_{\overline g} u - \frac{n}{2(n-1)}\Delta_{\overline g} u - \frac{n}{2(n-1)} u + O(t)\right)+O(t^4)\\ & = R_{g_0(t)} - \frac 12 t^2(\Delta_{\overline g}u +nu) +O(t^3) \end{align*} Here, I've used the fact that $\Gamma_{g_0(t)} = \Gamma_{\overline g} + O(t)$.