Typo or not in those GR notes? Given 
$$ds^2=-A(r)dt^2+B(r)dr^2+2C(r)drdt+D(r)r^2(d\theta^2+\sin^2d\phi^2),\tag{23.1}$$
we want to eliminate that cross term $2C(r)drdt$
Upon change of variable such that. This can happen upon
$$T(t,r)=t+\psi(r)\tag{23.2}$$ 
where $$dT^2=dt^2+\psi'^2dr^2+2\psi'drdt.$$and choosing $\psi$ to satisfy
$$\frac{d\psi(r)}{dr}=-\frac{C(r)}{A(r)},\tag{23.4}$$
says Blau in his lecture notes (http://www.blau.itp.unibe.ch/GRLecturenotes.html), page 481.
Now, let us try to work backwards and see if this helps us eliminate the cross term $2C(r)dtdr$.
We plug $\psi(r)/dr=\psi'=-C/A$ in $dT^2$ above. We get $$dT^2=dt^2+C^2/A^2-2C/Adrdt.$$
Plugging the latter result in $ds^2$, we get
$$-Adt^2-C^2/Adr^2+2Cdrdt+Bdr^2+Dr^2d\Omega^2.$$ 
And thus the cross term is NOT eliminated but rather summed up.
As I am trying to do the details to see if this works, turns out it does not work unless 
$$\frac{d\psi(r)}{dr}=\frac{C(r)}{A(r)},\hspace{1cm}\text{ dropped the minus sign}.$$
Am I on the right track or am I mistaken here? I am not sure if this is a typo or not.
 A: The book is right. Sticking to the time-radius sector, we are looking at the transformation
\begin{align}
t & \to T = t + \psi, \\
r & \to r,
\end{align}
where $\psi$ is a function of only $r$. In the new coordinates, we have
\begin{align}
\require{cancel}
g_{Tr} & = \frac{\partial x^\mu}{\partial T} \frac{\partial x^\nu}{\partial r} g_{\mu\nu} \\
& = \cancelto{1}{\frac{\partial t}{\partial T}} \left(\cancelto{-\psi'}{\frac{\partial t}{\partial r}} g_{tt} + \cancelto{1}{\frac{\partial r}{\partial r}} g_{tr}\right) + \cancelto{0}{\frac{\partial r}{\partial T}} \left(\frac{\partial t}{\partial r} g_{rt} + \frac{\partial r}{\partial r} g_{rr}\right) \\
& = -\psi' g_{tt} + g_{tr}.
\end{align}
For $g_{Tr}$ to vanish, it is necessary and sufficient to choose
$$ \psi' = \frac{g_{tr}}{g_{tt}} = -\frac{C}{A}. $$
A: The book is right, in fact, but we can be sure by checking a simple example. Let's take all your functions ($A, B, C, D$) equal to $1$, and ignore the angular part:
$$ds^2 = -dt^2 + dr^2 +2\,dt\,dr$$
Now the question is whether $\psi = r$ (your version) or $\psi = -r$ (the book's version). Let's say $\psi = \epsilon\, r$, with $\epsilon = \pm 1$. Replacing:
$$ds^2 = -dT^2 -2\epsilon\, dr^2 + 2(\epsilon+1)dT\,dr$$
So we need $\epsilon =-1$, that is, $d \psi / dr = -C/A$.
Edit: You seem to be having trouble with a particular thing. The original coordinates are $(t,r)$. Therefore, finding $dT^2$ in terms of $dt$ and $dr$ is not very useful, since what we want to do is to find $dt^2$ in terms of $dT$ and $dr$! The relation is $dt = dT-\psi' dr$ and so $dt^2 = dT^2 -2\psi' dt\,dr +\psi'^2 dr^2$, which you should be plugging into your original expression for $ds^2$ to find the new metric written in terms of $T$ and $r$.
