Singular jacobian matrix? I have a series of questions, in various degrees of befuddled muddledness (and they are related to my previous questions: this and this)

First question: how do I do a change of variable if the determinant of the jacobian is singular?

The setting for this question is as follows: I have one $n$-dimensional standard gaussian random variable $u \sim N(0,I)$ and a fixed $v \in \mathbb{R}^n$. Then I define the random variable
$$z = u - \frac{u^Tv}{v^Tv}v$$
and I'd like to derive a density for $z$. So the Jacobian:
$$\frac{dz}{du} = I - \frac{vv^T}{v^Tv}$$
which turns out to be singular and so $|\frac{dz}{du}|=0$. Does this mean trying to do a change of variable is fundamentally wrong here? Or is there a way to do this?

Second question: Aside from the Jacobian, I'm not sure how to change a standard normal distribution on $u$ to a distribution on $z$. So if the density on $u$ is
$$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2)$$
is there an inverse function $z^{-1}$ such that $z^{-1}(z) = u$? Then (I think)
$$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2) du = \frac{1}{\sqrt{2\pi}}\exp(-z^{-T}z^{-1}/2) \left|\frac{dz}{du}\right| du$$
So, is there such a $z^{-1}$? And if there is, what is it?

Third question: ultimately, I'm trying to answer this question. Am I going about this the right way by asking the two questions above? (The person who replied to my question there says something about Haar measure which I'd never heard of before so it's not enlightening to me as a proof.)
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 ;)
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 ;)