Let $\mathcal G$ and $\mathcal H$ be independent centered Gaussian processes with covariance function $(s,t)\mapsto \min(s,t) - st$. I want to compute the variance of $$X:=\left(-\int_0^1\mathcal G(s)\,\mathrm ds + \int_0^1\mathcal H(s)\,\mathrm ds\right).$$
Since \begin{align*} X^2 &= \left(-\int_0^1\mathcal G(s)\,\mathrm ds + \int_0^1\mathcal H(s)\,\mathrm ds\right)\left(-\int_0^1\mathcal G(s)\,\mathrm ds + \int_0^1\mathcal H(s)\,\mathrm ds\right) \\ &= \int_0^1\int_0^1\mathcal G(s)\mathcal G(t)\,\mathrm ds\,\mathrm dt + \int_0^1\int_0^1\mathcal H(s)\mathcal H(t)\,\mathrm ds\,\mathrm dt - \int_0^1\int_0^1\mathcal G(s)\mathcal H(t)\,\mathrm ds\,\mathrm dt - \int_0^1\int_0^1\mathcal H(s)\mathcal G(t)\,\mathrm ds\,\mathrm dt. \end{align*} Using Fubini, I can interchange expectation and integration operations. Since the Gaussian processes are centered, I obtain $\operatorname E\mathcal G(s)\mathcal G(t) = \min(s,t) - st$, $\operatorname E\mathcal H(s)\mathcal H(t) = \min(s,t)-st$, $\operatorname E\mathcal H(s)\mathcal G(t) = 0$, and $\operatorname E\mathcal H(t)\mathcal G(s) = 0$. Hence, \begin{align*} \operatorname EX^2 &= \int_0^1\int_0^1(\min(s,t)-st)\,\mathrm ds\,\mathrm dt + \int_0^1\int_0^1(\min(s,t)-st)\,\mathrm ds\,\mathrm dt - 0 \\ &= 2\int_0^1\left(\int_0^ts\,\mathrm ds + \int_t^1t\,\mathrm ds\right)\,\mathrm dt - 2\int_0^1\int_0^1st\,\mathrm ds\,\mathrm dt \\ &= \int_0^1 (2t-t^2)\,\mathrm dt - \frac 12 \\ &= \frac 2{12}. \end{align*} However, the solution is supposed to be $\frac 1{12}$. But I cannot spot my mistake....
Let \begin{equation*} \hat{F}_X^{(m)}=n^{-1}\sum_{i=1}^{m}\boldsymbol{1}_{X_i\le t}, \quad \hat{F}_X^{(n)}=n^{-1}\sum_{j=1}^{n}\boldsymbol{1}_{Y_j\le t}. \end{equation*} In above, by Functional Delta Method, you derive that \begin{align*} &\sqrt{n}[\phi(\hat{F}_X^{(n)},\hat{F}_Y^{(n)})-\phi(F_X,F_Y)]\\ &\quad \stackrel{\text{in dist.}}{\longrightarrow} -\int_0^1\mathcal{H}\,\mathrm{d}F_X+\int_0^1\mathcal{G}\,\mathrm{d}F_Y \stackrel{d}{=} N(0,\tfrac16). \end{align*} This fact consist also with the (40) and (42) in the notes you mentioned. It suffice notice that, here $N=2n$ and $n=n_1=n_2$, i.e. \begin{equation*} \sqrt{N}(U-1/2)\approx N(0,N^2/(12n^2)),\quad \sqrt{n} (U-1/2)\approx N(0,1/6)). \end{equation*}