Consider the region $D$ bounded by the straight lines $x=0$, $y=0$, and $x+y=1$. Clearly, the region is a triangle in the first quadrant. I want to transform this region using the transformation $x=u(1-v)$ and $y=uv$ for the purpose of evaluating a double integral.
As we can see,
$x=0 \implies u=0, v=1$
$y=0 \implies u=0, v=0$
$x+y=1 \implies u=1$
Does this mean that the straight line boundary $x=0$ transformed into a pair of straight lines ($u=0$ and $v=1$) on the $uv$ plane?
Your transformation is not a map $(x,y)\mapsto(u,v).$ It is a map $(u,v)\mapsto(x,y).$ So, your question makes little sense.
Even when written $x+y=u,y=uv,$ the transformation is not a map from the $xy$ to the $uv$ plane. It is undefined on the straight line $x=0$ you are interested in.
The only thing we can say is that your open triangle is transformed into the open square $(0,1)^2,$ because $$\begin{align}(u(1-v)>0\land uv>0\land u<1)&\iff\\(v(1-v)>0\land uv>0\land u<1)&\iff\\0<v<1\land0<u<1. \end{align}$$ This proves the last line of the solution in your attached image.