I'm starting with the argument provided by @Ross Millikan. Let $A=(0,0),\ B=(1,0),\ C=(0,1)$. Then the point chosen according to the given equation is $P=(X,Y)=(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$. Now clearly, $0\leq X,Y \leq 1$ and $X+Y\leq \sqrt{r_1}\leq 1$. Now the problem is to show that $\mathbb{P}(X\leq x, Y\leq y)=2xy,\ \forall 0\leq x,y\leq 1$ with $x+y\leq 1$. Now, \begin{equation*}
\begin{split}
\mathbb{P}(X\leq x, Y\leq y)=& \mathbb{P}(\sqrt{r_1}(1-r_2)\leq x, r_2\sqrt{r_1}\leq y)\\
\ =&\int_{0}^1 \mathbb{P}(\sqrt{r}(1-r_2)\leq x, r_2\sqrt{r}\leq y|r_1=r)f_{r_1}(r)dr\\
\ =&\int_{0}^1 \mathbb{P}(1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}})I_{[0,1]}(r)dr\ \mbox{(Since, $r_1, r_2$ are i.i.d $\mathcal{U}[0,1]$})\\
\end{split}
\end{equation*}
Now to find the region of integration we note that $$1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}}\ \Rightarrow\ 0\leq r\leq(x+y)^2$$ Also, if $x\leq y$ then
$$ r\in (0,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\
r\in (x^2,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\
r\in (y^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$
and if $y\leq x$ then
$$ r\in (0,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\
r\in (y^2,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\
r\in (x^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$

Then if $x\leq 1$ the integral becomes $$\int_{0}^{x^2}1 dr+\int_{x^2}^{y^2}\frac{x}{\sqrt{r}} dr+ \int_{y^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$
Similarly, if $y\leq x$ the integral becomes $$\int_{0}^{y^2}1 dr+\int_{y^2}^{x^2}\frac{y}{\sqrt{r}} dr+ \int_{x^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$ Hence the point $P$ is uniformly distributed on the surface of the triangle $ABC$. $\hspace{3cm}\ \Box$