I understand that the question is about the map
$$T:\quad(u,v)\mapsto (x,y)=T(u,v):=(u^2-v^2,2uv)\ ,$$
restricted to the given square $Q$.
${\bf 1\ }$ Assume we have two points $(u_i, v_i)\in Q$ with $T(u_1,v_1)=T(u_2,v_2)=:(x,y)$. Then
$$x^2+y^2=(u_i^2-v_i^2)^2+4u_i^2v_i^2=(u_i^2+v_i^2)^2\qquad(i=1, 2)\ .$$
This allows to conclude that $u_1^2+v_1^2=u_2^2+v_2^2$, and together with $u_1^2-v_1^2=x=u_2^2-v_2^2$ it follows that $u_1^2=u_2^2$, $v_1^2=v_2^2$. Since $Q$ is in the first quadrant we may take square roots here, and obtain $(u_1,v_1)=(u_2,v_2)$.
${\bf 2\ }$ The following picture incorporates also part ${\bf 4}$:

${\bf 3\ }$ The Jacobian is
$$dT(u,v)=\left[\matrix{2u&-2v\cr 2v&2u\cr}\right]\ ,$$
hence
$$dT(1,1)=\left[\matrix{2&-2\cr 2&2\cr}\right]=\sqrt{8}\left[\matrix{{1\over\sqrt{2}}&{-1\over\sqrt{2}}\cr {1\over\sqrt{2}}&{1\over\sqrt{2}}\cr}\right]\ .$$
The last matrix describes a counterclockwise rotation by $45^\circ$, followed by a scaling. Therefore it transforms squares into squares.
${\bf 4\ }$ The map $$(u,v)\mapsto T(1,1)+dT(1,1).(u-1,v-1)$$
is an approximation to $T$ in the neighborhood of $(1,1)$ and maps $Q$ to some euclidean square $Q'$ which "approximates" the actual image $T(Q)$.