EDIT
I will start my question with an example. Let us suppose that $f_{0}:[x_{1},x_{2}]\to[0,1]$ is a continuous function such that $0 < x_{1} < x_{2} < 1$ $|f^{-1}_{0}(\{y\})| < \infty$ for every $y\in[0,1]$. Since the domain is compact, it attains a maximum and a minimum at $x_{\max}$ and $x_{\min}$ respectively. Consequently, we can transform it into a surjective function according to the expression: \begin{align*} f(x) = \frac{f_{0}(x) - f_{0}(x_{\min})}{f_{0}(x_{\max}) - f_{0}(x_{\min})} \end{align*}
Can we always transform a continuous function $f_{0}:[x_{1},x_{2}]\times[y_{1},y_{2}]\to[0,1]\times[0,1]$, where $0 < x_{1} < x_{2} < 1$ and $0 < y_{1} < y_{2} < 1$, into a surjective function through translations and scalings applied to each coordinate function? If it is not possible in general, which restrictions should I impose in order to turn it feasible? Here, I assume that $|f^{-1}_{0}(\{(w,z)\})|<\infty$ for every $(w,z)\in[0,1]\times[0,1]$.
This is not homework. Any help is appreciated.
Rectangles are invariant under translation and scaling of each coordinate, so if $f_0$ has an image which is not a rectangle, then none of your transformations of it will have a rectangle as their image either, and in particular its image cannot be $[0,1] \times [0,1]$.
For instance, take $[x_1, x_2] = [\frac{1}{4}, \frac{1}{2}]$ and $[y_1, y_2] = [0, 2\pi]$, and $$f_0(x,y) = (f_1(x,y), f_2(x,y)) = (x \cos y + \frac{1}{2}, x \sin y + \frac{1}{2}).$$ The image of $f_0$ is an annulus, and any function of the form $a_1 f_1(x,y) + b_1, a_2 f_2(x,y) + b_2$ will have an image which is an elliptical annulus of some kind, or perhaps a line segment or a point if one or both of the $a_i$ are zero. But it can never be a square.