I think for the second question... If $f$ is the density of v.a. $u$ with
$$f(u\;\vert\;\mu={\bf 0},\sigma={\bf I})$$
according to the transformation $z$, then
$$f(z\;\vert\;\tilde{\mu},\tilde{\sigma})=f\left(u(z)\;\vert\;\mu={\bf 0},\sigma={\bf I}\right)\left|\frac{\partial u}{\partial z}\right|$$
(respect to $dz$!!!)
P.D.: Excuse my English ;)
We are talking here about a theorem covered in many Advanced Calculus texts. But I have not found a proof on the web. Wikipedia treats "functional dependence" in a purely semantic way.
Assume that both $f$ and $g$ are defined in a neighborhood $U$ of $(0,0)\in{\mathbb R}^2$, and that $$g(0,0)=0 , \qquad g_{.2}(0,0)\ne0\ .$$
Claim. If $$\nabla f(x,y)\wedge \nabla g(x,y)=0\qquad\forall (x,y)\in U\tag{1}$$ then there is a $C^1$-function $t\mapsto h(t)$, defined in a neighborhood of $t=0$, such that
$$f(x,y)=h\bigl(g(x,y)\bigr)\qquad\forall\ (x,y)\in U\ .$$
Proof. Consider the auxiliary function
$$F(x,y,t):=g(x,y)-t\ .$$
As $F(0,0,0)=0$ and $F_y(0,0,0)=g_{.2}(0,0)\ne0$ the implicit function theorem allows to solve $F(x,y,t)=0$ in a neighborhood of $(0,0,0)$ for the variable $y$: There is a $C^1$-function $\psi$, defined in a neighborhood of $(0,0)$, such that $g(x,y)=t$ is equivalent with $y=\psi(x,t)$. It follows that $g\bigl(x,\psi(x,t)\bigr)\equiv t$, so that
$$g_{.1}\bigl(x,\psi(x,t)\bigr)+g_{.2}\bigl(x,\psi(x,t)\bigr)\psi_x(x,t)\equiv0\ ,$$
or
$$\psi_x(x,t)=-{g_{.1}\bigl(x,\psi(x,t)\bigr)\over g_{.2}\bigl(x,\psi(x,t)\bigr)}\ .\tag{2}$$
Now define
$$\tilde h(x,t):=f\bigl(x,\psi(x,t)\bigr)\ .\tag{3}$$
I claim that the dependence of $\tilde h$ on $x$ is only apparent. To prove this compute
$$\tilde h_x=f_{.1}\bigl(x,\psi(x,t)\bigr)+f_{.2}\bigl(x,\psi(x,t)\bigr)\psi_x(x,t)\equiv0\ ,$$
using $(2)$ and $(1)$. We therefore may replace $(3)$ by
$$h(t):=f\bigl(x,\psi(x,t)\bigr)\ .\tag{4}$$
Consider now an arbitrary point $(x,y)\in U$, and put $g(x,y)=:t$. Then $y=\psi(x,t)$, and $(4)$ gives
$$f(x,y)=f\bigl(x,\psi(x,t)\bigr)=h(t)=h\bigl(g(x,y)\bigr)\ .\qquad\qquad\square$$
Best Answer
Let $f_1$, $\ldots$, $f_{k+1}$ such that the jacobian of $(f_1, \ldots, f_k)$ has maximal rank $k$, but the jacobian of $(f_1, \ldots,f_{k+1})$ has constant rank $k$. Then $f_{k+1}$ is a function of $f_1$, $\ldots$, $f_k$.
Indeed, any locus $(f_1,\ldots, f_k)=$ const is a manifold of codimension $k$. Because of the condition on the jacobian of $(f_1, \ldots,f_{k+1})$ the function $f_{k+1}$ is constant on such a locus. It follows that $f_{k+1}= F(f_1, \ldots, f_k)$, for $F$ function of $k$ variables. This function will be smooth, since the map $(f_1, \ldots, f_k)$ is a submersion.
$\bf{Added:}$ Let's explain why if $f_1$, $\ldots$, $f_k$ have jacobian of maximal rank $k$, and $\operatorname{grad}f$ is linearly dependent on $\operatorname{grad}f_1$, $\ldots$, $\operatorname{grad}f_k$, then $f$ is locally constant on any submanifold of the form $f_i = c_i$, $i=1, k$. Indeed, consider a smooth path $t \mapsto \phi(t)$ on such a submanifold. Then at every $t$ we have $$\langle \operatorname{grad}f_i\ _ {\phi(t)}, \phi'(t)\rangle = 0$$ and by the linear dependence of the gradients we get $$\langle \operatorname{grad}f _ {\phi(t)}, \phi'(t)\rangle = 0$$ and so the function $f(\phi(t))$ is constant in $t$.