Let$ D(z, r) \subset \Bbb R^2$ be disc and $Q(a,\rho)= (a_1-\rho,a_1+\rho)\times(a_2-\rho,a_2+\rho)\subset \Bbb R^2 $ be a square such that $$|D(z,r)|= |Q(a,\rho)|$$ Can we construct explicitly a bijection a continuous $\phi: D(z, r) \to Q(a, \rho)$?
More generally, is there a general explicit formula $\phi(x,y)$ on $\Bbb R^2$ such that: the image of a Disc is a square of the same volume?
That is, for given a disc $ D(z, r) \subset \Bbb R^2$,there is square $Q(a,\rho)= (a_1-\rho,a_1+\rho)\times(a_2-\rho,a_2+\rho)\subset \Bbb R^2 $ such that $\phi: D(z, r) \to Q(a, \rho)$ is a bijection and
$$|D(z,r)|= |Q(a,\rho)|?$$
A sketch of an answer.
Let $z\in\mathbb{R}^2$ be arbitrary and let $C(z,r)$ be a circle around $z$ with radius $r>0$. Define the function $\phi:\mathbb{R}^2\to\mathbb{R}^2$ by projecting any circle $C(z,r)$ onto the square centered at $z$ with side length $r\sqrt{\pi}$. This is clearly bijective and continuous and maps any circle centered at $z$ to a square of the same area.
Now for the more interesting question, which is what I assume you meant by part $2$ of your post. Does there exist a continuous bijection $\phi:\mathbb{R}^2\to\mathbb{R}^2$ that maps any disc in the plane to a square with the same area?
Claim: Such a function cannot exist.
Assume the contrary and let $\phi$ be such a function. Fix some $\delta>0$ and consider the discs $D$ and $D'$ centered at $0$ with radii $1$ and $1-\delta$ respectively and let $C$ and $C'$ be their boundary circles. For each $0\leq \theta<2\pi$ consider the disc $D_\theta$ of radius $\frac{\delta}{2}$ tangent to both $C$ and $C'$ such that the intersection point with $C$ has argument $\theta$ and let $C_\theta$ be its boundary circle.
Now define $S,S',S_\theta$ to be the corresponding images through $\phi$ of $D,D',D_\theta$ and let $\Sigma,\Sigma',\Sigma_\theta$ be their respective boundaries. Since $\phi$ is continuous, it is not hard to see that $\Sigma=\phi(C)$ and similar for $\Sigma'$ and $\Sigma_\theta$. Also, it is clear that $S',S_\theta\subset S$ for all $\theta$. Additionally, since the areas are preserved, we must have that the side length of $S$ is $\sqrt{\pi}$, that of $S'$ is $(1-\delta)\sqrt{\pi}$ and that of $S_\theta$ is $\frac{\delta\sqrt{\pi}}{2}$ (for any $\theta$).
Now, since $\phi$ is bijective, we must have that, for each $\theta$, $\Sigma_\theta\cap\Sigma$ has exactly one element and so does $\Sigma_\theta\cap\Sigma'$. Pick an arbitrary point $P\in\Sigma'$. The previous statement implies $\delta\frac{\sqrt{\pi}}{2}<dist(P,\Sigma)<\delta\frac{\sqrt{\pi}}{\sqrt{2}}$ since we can construct the tangent disc $D_{\arg{\phi^{-1}(P)}}$ which will have radius $\delta\frac{1}{2}$ and therefore we can construct the corresponding tangent square $\Sigma_{\arg{\phi^{-1}(P)}}$ with radius $\delta\frac{\sqrt{\pi}}{2}$.
However, it is not difficult to see that no matter what the relative orientation of $\Sigma'$ is inside of $\Sigma$, there will always exist some point $P\in\Sigma'$ such that $dist(P,\Sigma)\leq \delta\frac{\sqrt{\pi}}{2}$ as the side lengths of $\Sigma$ and $\Sigma'$ were $\sqrt{\pi}$ and $(1-\delta)\sqrt{\pi}$ respectively and since $S'\subset S$, which gives us a contradiction.
I suspect there might be a more 'measure-theoretic' approach to proving the Claim since it is not difficult to see that if such a $\phi$ exists, it will preserve the area of at least any shape with a piece-wise smooth boundary (by approximating the area from above and below with unions of disjoint discs to arbitrary precision).