Let $T$ be the triangle bounded by
$$
x=0,\\
y=0,\\
x+y=1
$$
Is there such a change of variables $$u=f_1(x,y),\\v=f_2(x,y)$$ (or a sequence of such changes) that transforms this triangle into the square $u=0,u=1, v=0,v=1$?
I have been trying to find these transformations, but I have not been able to do so yet.
An intuitive solution would be to imagine the triangle as elastic, and pull the $(1/2,1/2)$ point to $(1,1)$. For every original point $(x,y)$ you calculate the polar angle $\theta$, you calculate the intersection of the line at angle $\theta$ with the $x+y+=1$ line, you calculate the intersection of the radial line with $x=1$ and $y=1$ lines, and scale the original $(x,y)$ by the ratio of the lengths to $x=1$ (or $y=1$, as appropriate) and length to $x+y=1$. Can you take it from here?
EDIT:
Let the intersection of $(0,0)\to(x,y)$ line be $(x_s,y_s)$, the intersection with $x=1$ be $(1,y_1)$, and the intersection with $y=1$ be $(x_1,1)$. To calculate these points, we just write the equations: $$\frac xy=\frac{x_s}{y_s}=\frac{x_1}1=\frac{1}{y_1}$$ This allow us to immediately say $$x_1=\frac xy\\y_1=\frac yx$$ For $x_s,y_s$ we also need to use $$x_s+y_s=1$$ Then $$y_s=\frac yx x_s\\x_s\left(1+\frac yx\right)=1\\x_s=\frac{x}{x+y}\\y_s=\frac{y}{x+y}$$ With these, the length of $(0,0)\to(x_s,y_s)$ is $$L_s=\frac{\sqrt{x^2+y^2}}{x+y}$$ The length of $(0,0)\to(1,y_1)$ is $$L_{y_1}=\frac{\sqrt{x^2+y^2}}{x}$$ and similarly $$L_{x_1}=\frac{\sqrt{x^2+y^2}}{y}$$ So for points with $x\gt y$ we scale both $x$ and $y$ by $$\frac{L_{y_1}}{L_s}=\frac{x+y}x$$ For $y\gt x$ the scale factor is $$\frac{L_{x_1}}{L_s}=\frac{x+y}y$$ Therefore $$u=x\frac{x+y}{\max(x,y)}\\v=y\frac{x+y}{\max(x,y)}$$