In this previous question: Transported Metric, I wanted to transport the Euclidean metric $ds^2=dx^2+dy^2$ into the first quadrant of the $(u,v)$-plane, such that for all curves $\gamma$ in the $(x,y)$-plane we have $L_{uv}\bigl(f(\gamma)\bigr)=L_{xy}(\gamma)$.
I used a map $f:\Bbb R^2\to\Bbb R^{2}$ with $f(x,y)=(e^x,e^y).$ Then (with the help of @ChristianBlatter) the transported metric is:
$$ds^2={1\over u^2}du^2+{1\over v^2}dv^2\ .$$
I noticed that $f$ maps the third quadrant of the $(x,y)$-plane to $(0,1)^2$ in the $(u,v)$-plane. So I decided to map all the quadrants of the $(x,y)$-plane into $(0,1)^2$ of the $(u,v)$-plane and write down the transported metric.
Since all the quadrants of the $(x,y)$-plane are represented in $(0,1)^2$ of the $(u,v)$-plane, then I think it is isomorphic to $\Bbb R^2.$
So the four maps I used were, $f_n:\Bbb R^2 \to \Bbb R^2$ for $n=1,2,3,4$ with:
$$ f_1(x,y)=(-e^{-x},-e^{-y})$$
$$ f_2(x,y)=(e^x,-e^{-y}) $$
$$ f_3(x,y)=(e^x,e^y) $$
$$ f_4(x,y)=(-e^{-x},e^y) $$
The subscripts represent the quadrant that each map "acts on." So $f_3$ for example, maps points from the third quadrant to $(0,1)^2.$ So I'll say that $f_3$ "acts on" points in the third quadrant.
Then I translated all the image spaces for $f_n$ to $(0,1)^2$ of the $(u,v)$-plane. Now the quadrants all overlap so I don't know how to define the metric.
What is the transported metric in $(0,1)^2$ of the $(u,v)$-plane?
Reading this and your previous post, I have the feeling that you have some misunderstanding of "transported metrics".
In general, let $f :X\to Y$ be a smooth mapping and $h$ is a metric on $Y$. Then one can pullback the metric $h$ to form a new symmetric two tensors on $X$, given by
$$ g_x( X, Y):= h_{f(x)} (df_x X , df_x Y),$$ for all $x\in M$ and $X, Y \in T_xM$. Since $h$ is a metric, $g$ is non-negative definite. If it is also positive definite (this is true if and only if $df_x$ is an immersion), then $g$ is called the pullback metric.
So pullback metric on the target to get a metric on the domain. Not the other way round.
In your previous post, this does not emerge since $f : \mathbb R^2 \to \mathbb R_+^2$, $f(x, y) = (e^x, e^y)$ is a diffeomorphism with inverse $f^{-1}(u, v) = (\log u, \log v)$. Indeed now one can pullback the euclidean metric $dx^2 + dy^2$ to $\mathbb R_+^2$ via $f^{-1}$: which is
\begin{align} (f^{-1})^* (dx^2 + dy^2) &=( d((f^{-1})^*x))^2 + ( d((f^{-1})^*y))^2 \\ &= (d (x\circ f^{-1}))^2 +(d (y\circ f^{-1}))^2 \\ &= (d \log u)^2 +(d\log v)^2 \\ &= \left( \frac{1}{u} du\right)^2 + \left( \frac{1}{v} dv\right)^2 \\ &= \frac{1}{u^2} du^2 + \frac{1}{v^2} dv^2. \end{align}
This agrees with the one given by the answer in the previous post.
So in general, you cannot pushforward a Riemannian metric. You can only pullback via an immersion.
Now go back to your question: You want a Riemannian metric on $(0,1)^2$ so that $$\tag{1} L_{uv} (f(\gamma)) = L_{xy}(\gamma)$$ this cannot be done and the reason is precisely that $f$ is not one to one (so not invertible): For example, both the straight lines $$\gamma_1 = \{1\} \times [-1, 1], \ \ \ \gamma_2 \{1\} \times [-1, 0]$$ satisfy $f(\gamma_1) = f(\gamma_2)$ but $L_{xy} (\gamma_1) \neq L_{xy} (\gamma_2)$. Thus (1) is impossible.
You could do something else: for any curve $\eta$ on $(0,1)^2$, one considers the preimage $f^{-1}(\eta)$, which is a disjoint union of four curves $$ f^{-1} \eta = \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4,$$ where $\gamma_i$ lies in the $i$-th quadrant for $i=1, 2, 3, 4$. From your construction of $f_i$, we indeed have $$ L_{xy}(\gamma_i) = L_{xy}(\gamma_j).$$
Then
$$ L_{xy} (f^{-1}(\eta)) = 4L_{xy} (\gamma_1).$$
So if we define the Riemannian metric $$ 4 \left( \frac{1}{u^2} du^2 + \frac{1}{v^2} dv^2\right)$$ on $(0,1)^2$, then $$\tag{2} L_{xy} (f^{-1}(\eta)) = L_{uv} (\eta)$$ for all curves $\eta$ on $(0,1)^2$. (2) is not quite what you want, which is (1).