Your notion of algebraically independent equations is not one that I am familiar with. Using your example, I would guess that the definition is that $f_1(x_1, \ldots, x_n)$, $f_2(x_1, \ldots, x_n)$, ..., $f_m(x_1, \ldots, x_n)$ are "algebraically independent as equations" if, given the knowledge of the values of any $m-1$ of the $f_i$, the remaining $f$ is still free to assume any value. I'm going to term this "logical independence". I have not seen this condition before, but I'll try to make some useful remarks about it.
Claim The polynomials $f_1$, $f_2$, ..., $f_m$ are logically independent iff the map
$$F:(x_1, \ldots, x_n) \mapsto (f_1(x), \ldots, f_m(x))$$
from $k^n \to k^m$ is surjective.
Proof: It is clear that, if $F$ is surjective, then the $f_i$ are algebraically independent. Conversely, suppose that the $f_i$ are logically independent. Let $(x_1, \ldots, x_m)$ be an arbitrary point in $k^m$.
We will show, by induction on $i$, that there is a point $(x_1, x_2, \ldots, x_i, y_{i+1}, \ldots, y_m)$ in the image of $F$. The base case, $i=0$, is vacuous. Suppose that $(x_1, \ldots, x_i, y_{i+1}, \ldots, y_m)$ is in the image of $F$. Since the $f_i$ are logically independent, the equations $f_1=x_1$, $f_2=x_2$, ..., $f_i=x_i$ does not let us deduce anything about $f_{i+1}$. In particular, it does not let us deduce $f_{i+1} \neq x_{i+1}$. So there is some $z$ with $f_1(z) = x_1$, ..., $f_i(z) = x_i$ and $f_{i+1}(z)=x_{i+1}$. This completes the induction. $\square$.
As your example demonstrates, logical independence is stronger than algebraic independence: $f_1$, ..., $f_m$ are algebraically independent (over an infinite field) if and only if there is a nonzero polynomial $P$ which vanishes on the image of $F$. A simpler example is
$$x \ \mbox{and} \ xy.$$
When $k$ is not algebraically closed, logical independence seems to be a very hard notion to work with. For example,
$$x^{37}+y^{37}+z^{37} \ \mbox{and} \ xyz$$
are logically dependent over $\mathbb{Q}$ by the work of Taylor and Wiles. It would probably be easy to find equations where it is still an open problem whether they are logically independent.
When $k$ is algebraically closed, this is a reasonable notion, but I've never seen a more concise name for it than "the map $(f_1, \ldots, f_m)$ is surjective".
When $k$ is algebraically closed and $m=n$, we have the following result: If $k[x_1, \ldots, x_n]$ is a finite $k[f_1, \ldots, f_n]$-module, this implies that the $f_i$ are logically independent. The adjective for this is "the map $F$ is finite". "Finite" is strictly stronger than surjective. For example, $(x,y) \mapsto (x+xy, x^2 y)$ is surjective over $\mathbb{C}$ (exercise!) but I claim that $\mathbb{C}[x+xy, x^2 y]$ is not a finite $\mathbb{C}[x,y]$ module. (Harder exercise. Hint: Show that $\mathbb{C}[x,xy]$ IS finite over $\mathbb{C}[x+xy,xy^2]$.)
I don't know of a good generalization of "finite" to the case $m \neq n$.
