Intuition behind functional dependence

Functional dependence of $k$ given functions $F_i:\>\Omega\to{\mathbb R}$ $\>(1\leq i\leq k)$ with common domain $\Omega\subset{\mathbb R}^n$ means, intuitively, that there is a nontrivial function $$g:\quad{\mathbb R}^k\to{\mathbb R},\qquad y\mapsto g(y)$$ such that $$g\bigl(F_1(x),F_2(x),\ldots, F_k(x)\bigr)\equiv 0\qquad \forall x\in\Omega\ .\tag{1}$$ "Nontrivial" for $g$ means that $\nabla g(y)\ne0$ for all $y\in{\mathbb R}^k$. Taking the derivative of $(1)$ we see that $$dg\bigl(F(x)\bigr)\cdot dF(x)\equiv 0\in{\cal L}({\mathbb R}^n,{\mathbb R}^k)\qquad\forall x\in\Omega\ .$$ In terms of matrices this says that the rows of the matrix $\bigl[dF(x)\bigr]$ are linearly dependent with coefficients given by $\nabla g\bigl(F(x)\bigr)\ne0$, for each $x\in\Omega$.

Now the rows of the matrix $\bigl[dF(x)\bigr]$ are nothing else but the gradients $\nabla F_i(x)$. Therefore functional dependence of the $F_i$ in the above sense implies that the gradients $\nabla F_i(x)$ are linearly dependent, at each point $x\in\Omega$.

In your definition of "functional independence" you don't want even a hint of such a thing. Therefore you insist that at all points $x\in\Omega$ the $k$ gradients $\nabla F_i(x)$ should be linearly independent.

Functional independence between two functions means that the level set of each of the functions intersects transversely the level set of the other function. In the plane, this means that a function $f$ functionally independent of another function $g$ cannot be written as $f=F(g)$ where $F$ is another function, because in this case, the level sets would be the same curves, for both $f$ and $g$. Similarly, in $k$ dimensions, functional independence means that the intersection between the $k$ $(k-1)$-dimensional level sets of the $k$ functions is a point, not a line or a plane, i. e., the functions $F_1$, $F_2$,...,$F_k$ contains $k$ independent informations of your ambient space, and can be taken locally as coordinates.