I started with $\Bbb R^2$ and the Euclidean metric $ds^2=dx^2+dy^2.$ Then used a map $f:\Bbb R^2\to\Bbb R^{2+}$ with $f(x,y)=(e^x,e^y).$
How do you write down the new metric after the map?
I know that under the map, the origin $(0,0)\mapsto(1,1).$ So $(1,1)$ is the new origin, in the image of the map.
I searched in "An Introduction to Manifolds" by Tu, to try to figure out how to write down the new metric after the map. I searched in the section about "push-forwards" because I think that's what I need.
So from following my text supplemented with a wikipedia article, if the tangent space is defined with differentiable curves $g_n(x)=\pm\frac{n}{x}$ and $df_p:T\Bbb R^2\to T\Bbb R^{2+}$ then $df_p(g'_n(0))=(f \circ g_n)'(0).$ I'm stuck at this part.
Your question "How do you write down the new metric after the map?" is difficult to understand. It seems that you want 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 $L_{uv}\bigl(f(\gamma)\bigr)=L_{xy}(\gamma)$.
Any $ds^2$ in the $(u,v)$-plane will be of the form $$ds^2=E(u,v)du^2+2F(u,v)dudv+G(u,v)dv^2$$ with certain functions $E$, $F$, $G$. (This is the oldfashioned way to write the $g_{ik}$ tensor.) The length of a curve $$\alpha:\quad t\mapsto\bigl(u(t),v(t)\bigr)\qquad(a\leq t\leq b)$$ then computes to $$L_{uv}(\alpha)=\int_a^b\sqrt{E(\alpha(t))u'^2(t)+2F(\alpha(t))u'(t)v'(t)+G(\alpha(t))v'^2(t)}\>dt\ .\tag{1}$$ Now we want $$L_{uv}(\alpha)=L_{xy}\bigl(f^{-1}(\alpha)\bigr)\ .$$ As $$f^{-1}(\alpha):\quad t\mapsto \bigl(x(t),y(t)\bigr):=\bigl(\log u(t),\log v(t)\bigr)$$ this means that $$L_{uv}(\alpha)=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\>dt=\int_a^b\sqrt{{u'^2(t)\over u^2(t)}+{v'^2(t)\over v^2(t)}}\>dt\ ,\tag{2}$$ for any curve $\alpha$. Comparing $(2)$ with $(1)$ we see that for the given map $f$ we need $$E(u,v)={1\over u^2},\quad F(u,v)\equiv0,\quad G(u,v)={1\over v^2}\ ,$$ so that the $ds^2$ in the $(u,v)$-plane will be $$ds^2={1\over u^2}du^2+{1\over v^2}dv^2\ .$$