Since $a^2 +b^2 - (a-b)^2 = 2ab$ we get
\begin{align}
\int_{0}^{\xi}\frac{\ln(t)\ln(1+t)}{1-t}\, \mathrm{d}t & = \frac12\int_{0}^{\xi} \frac{\ln^2(t)}{1-t}\, \mathrm{d}t + \frac12\int_{0}^{\xi} \frac{\ln^2(1+t)}{1-t}\, \mathrm{d}\color{blue}{t} - \frac12\int_{0}^{\xi} \frac{\ln^2\left( \frac{t}{t+1}\right)}{1-t}\, \mathrm{d}\color{purple}{t}\\
& \overset{\substack{\color{blue}{t+1\to t} \\ \color{purple}{t/(t+1) \to t}}}{=} \frac12\int_{0}^{\xi} \frac{\ln^2(t)}{1-t}\, \mathrm{d}t + \frac12\int_{1}^{\xi +1} \frac{\ln^2(t)}{2-t}\, \mathrm{d}\color{blue}{t}- \frac12\int_{0}^{\frac{\xi}{\xi +1}} \ln^2\left(t\right)\left(\frac{1}{\frac12-t} - \frac{1}{1-t}\right)\, \mathrm{d}\color{purple}{t} \\
& = \frac12\int_{0}^{\xi} \frac{\ln^2(t)}{1-t}\, \mathrm{d}t - \frac12\int_{0}^{1} \frac{\ln^2(t)}{2-t}\, \mathrm{d}t + \frac12\int_{0}^{\xi +1} \frac{\ln^2(t)}{2-t}\, \mathrm{d}t- \frac12\int_{0}^{\frac{\xi}{\xi +1}} \frac{\ln^2\left(t\right)}{\frac12-t} \, \mathrm{d}t + \frac12\int_{0}^{\frac{\xi}{\xi +1}} \frac{\ln^2\left(t\right)}{1-t}\, \mathrm{d}t\tag{1}\\
\end{align}
This reduces the problem to evaluating
\begin{equation}
\int_{0}^{\alpha} \frac{\ln^2(t)}{\beta -t}\, \mathrm{d}t \overset{\color{blue}{t/\beta \to t}}{=} \int_{0}^{\frac{\alpha}{\beta}} \frac{\ln^2(\beta) + 2 \ln(\beta)\ln (t)+\ln^2(t)}{1-t}\, \mathrm{d}t \tag{2}
\end{equation}
Which in turn reduces the problem to evaluating $\int_{0}^{x} \frac{\ln^n(t)}{1-t}\, \mathrm{d}t$ for $n=0,1,2$.
The $n=0$ case is simply $\int_{0}^{x} \frac{\mathrm{d}t}{1-t} = - \ln(1-x)$. For $n=1$ we recall that the polylogatithm follows the recursion $\mathrm{Li}_{s}(z) = \int_{0}^{z} \frac{\mathrm{Li}_{s-1}(z)}{t}\, \mathrm{d}t$ and that $\mathrm{Li}_{1}(z) =- \ln(1-z)$. Thus
$$
\int_{0}^{x}\frac{\ln(t)}{1-t}\, \mathrm{d}t \overset{\color{blue}{1-t \to t}}{=} -\int_{1}^{1-x}\frac{\ln(1-t)}{t}\, \mathrm{d}t = \mathrm{Li}_{2} (1-x) - \mathrm{Li}_{2} (1) = \mathrm{Li}_{2} (1-x) - \zeta(2)
$$
since $ \mathrm{Li}_{s} (1) = \zeta(s)$ for $s>1$ is also a property of the polylogarithm function.
Lastly, for $n=2$ we get
\begin{align}
\int_{0}^{x} \frac{\ln^2(t)}{1-t}\, \mathrm{d}t &\overset{\text{IBP}}{=} -\ln^2(x)\ln(1-x) + 2 \int_{0}^{x} \ln(t)\frac{\ln(1-t)}{t}\, \mathrm{d}t\\
&\overset{\text{IBP}}{=} -\ln^2(x)\ln(1-x) - 2 \ln(x)\mathrm{Li}_2(x)+ 2 \int_{0}^{x} \frac{\mathrm{Li}_2(t)}{t}\, \mathrm{d}t\\
& = -\ln^2(x)\ln(1-x) - 2 \ln(x)\mathrm{Li}_2(x)+ 2 \mathrm{Li}_3(x)
\end{align}
Plugging the $n=0,1,2$ cases into $(2)$ gives
\begin{align}
\int_{0}^{\alpha} \frac{\ln^2(t)}{\beta -t}\, \mathrm{d}t& = -\ln^2(\beta)\ln\left(1- \frac{\alpha}{\beta} \right) +2 \ln(\beta)\mathrm{Li}_2\left(1- \frac{\alpha}{\beta} \right) -2\ln(\beta)\zeta(2) -\ln^2\left(\frac{\alpha}{\beta} \right)\ln\left(1- \frac{\alpha}{\beta} \right) - 2 \ln\left(\frac{\alpha}{\beta} \right)\mathrm{Li}_2\left(\frac{\alpha}{\beta} \right)+ 2 \mathrm{Li}_3\left(\frac{\alpha}{\beta} \right)
\end{align}
There's now a bit of simplification that can be done. Since \begin{align}
\mathrm{Li}_2(x) +\mathrm{Li}_2(1-x) & = \zeta(2)-\int_{0}^{x} \frac{\ln(1-t)}{t}- \frac{\ln(t)}{1-t}\mathrm{d}t \\
& = \zeta(2)-\int_{0}^{x} \left(\ln(t)\ln(1-t)\right)'\mathrm{d}t\\
& = \zeta(2) - \ln(x)\ln(1-x)
\end{align}
We get that
$$
\int_{0}^{\alpha} \frac{\ln^2(t)}{\beta -t}\, \mathrm{d}t = 2\mathrm{Li}_3\left(\frac{\alpha}{\beta} \right) - 2 \ln(\alpha) \mathrm{Li}_2\left(\frac{\alpha}{\beta} \right) -\ln^2(\alpha)\ln\left(1- \frac{\alpha}{\beta} \right)
$$
And using the previous equation in $(1)$ we get that
\begin{align}
\int_{0}^{\xi}\frac{\ln(t)\ln(1+t)}{1-t}\, \mathrm{d}t& = \text{Li}_3(\xi)+\text{Li}_3\left(\frac{\xi}{1+\xi}\right)-\text{Li}_3\left(\frac{2
\xi}{1+\xi}\right)+\text{Li}_3\left(\frac{1+\xi}{2}\right)-\text{Li}_2(\xi) \log
(\xi)-\text{Li}_2\left(\frac{\xi}{1+\xi}\right) \log
\left(\frac{\xi}{\xi+1}\right)+\text{Li}_2\left(\frac{2 \xi}{1+\xi}\right) \log
\left(\frac{\xi}{\xi+1}\right)-\text{Li}_2\left(\frac{1+\xi}{2}\right) \log
(\xi+1)-\frac{1}{2} \log (1-\xi) \log ^2(\xi)+\frac{1}{2} \log \left(\frac{1-\xi}{\xi+1}\right)
\log ^2\left(\frac{\xi}{\xi+1}\right)+\frac{1}{2} \log ^2\left(\frac{\xi}{\xi+1}\right) \log
(\xi+1)-\frac{1}{2} \log \left(\frac{1-\xi}{2}\right) \log ^2(\xi+1)+\frac{1}{24} \left(-21
\zeta (3)-4 \log ^3(2)+2 \pi ^2 \log (2)\right)
\end{align}
and plugging in $\xi = \frac13$ gives the evaluation of the desired integral as
$$
\int_{0}^{\frac13}\frac{\ln(t)\ln(1+t)}{1-t}\, \mathrm{d}t = \text{Li}_3\left(\frac{1}{4}\right)+\text{Li}_3\left(\frac{1}{3}\right)+\text{Li}_3\left(
\frac{2}{3}\right)+\frac{1}{2} \log \left(\frac{4}{3}\right) \left(\log ^2(4)-2
\text{Li}_2\left(\frac{2}{3}\right)\right)+\text{Li}_2\left(\frac{1}{4}\right) \log
(4)+\text{Li}_2\left(\frac{1}{3}\right) \log (3)-\frac{7 \zeta (3)}{4}+\frac{2 \log
^3(2)}{3}+\frac{1}{2} \log ^2\left(\frac{4}{3}\right) \log (3)+\frac{1}{2} \log
\left(\frac{3}{2}\right) \log ^2(3)-\frac{1}{2} \log (2) \log ^2(4)
$$
