The specific form of the Killing vectors depends on the parameterization of the group element, from the notation (and the results), one can deduce that Euler angle parameterization has been used:
$ g = exp(i\sigma_3 \psi) exp(i \sigma_1 \theta) exp(i \sigma_3 \phi)$
where the sigmas are the generators of rotations with respect to Cartesian axes in the three dimensional representation.
Also the two sets of killing vectors correspond to the left and right action of $SO(3)$ on itself which preserve the invariant metric. I'll describe to you the case of the left action for example.
The basic equation definining the lect killing vectors is
$ K_A^L g = \sigma_A g$ (for the right action $K_A^R g = g \sigma_A$, I'll skip from now the superscript understanding it is a left action).
$K_A$ is a differential operator:
$K_A = K_A^{\phi} \frac{\partial}{\partial \phi} +K_A^{\theta} \frac{\partial}{\partial \theta} +K_A^{\psi} \frac{\partial}{\partial \psi}$
For convenience we shall call $x^1 = \phi$, $x^2 = \theta$, $x^3 = \psi$,
So our task is to compute $K_A^j$
In order to do that,we remember that Maurer-Cartan one form $g^{-1} dg$ is a Lie algebra valued one form, i.e.,
$m = g^{-1} dg = i a_j^A \sigma_A dx^j$ (With summation convention)
Thus, the first task to be done is to explicitely compute the coefficients $ a_j^A$, this is done by computing the derivatives in the given parameterization.
If we contract this form with a killing vector, we obtain:
$<K_A, m> = i K_A^j a_j^B \sigma_B = g^{-1} K_A g =g^{-1} \sigma_A g$
Using the orthogonality relations
$tr(\sigma_A \sigma_B) = \delta_{AB}$
We obtain:
$K_A^j a_j^B = tr(\sigma_B g^{-1} \sigma_A g)$
Thus by solving this system of linear equations or equivalently, inverting the matrix A we get the formula for the Killing vectors components:
$K_A^j = (a^{-1})_B^j tr(\sigma_B g^{-1} \sigma_A g)$
in summary, one needs to compute the coefficient matrix of the Maurer-Cartan form and invert it and to compute the traces required by the last equation, then compute the Killing vector components by matrix multiplication.
Partial answer:
About the symmetry, the part : $-(1+r^2)d\tau^2+\frac{dr^2}{1+r^2}$ is just the metrics of AdS2, so there is a $SO(2,1)$ symmetry.
At fixed $\theta$, we have this $SO(2,1) \sim SL(2,R)$ symmetry plus the $U(1)$ symmetry corresponding to the invariance by $\phi$ translation.
The $\theta$ parameters are playing the role of geometrical terms, but do not change the nature of the symmetry.
Ref, Chapter 2, page 3
[EDIT]
Correcting OP calculus:
$\mathcal{L}_\xi g_{\tau\tau}=\xi^\sigma\partial_\sigma g_{\tau\tau}+g_{\sigma\tau}\partial_\tau\xi^\sigma+g_{\tau\sigma}\partial_\tau\xi^\sigma \tag{1}$
$g_{\tau\tau}$ depends only on $r$, and the only non-null metrics $g_{\sigma\tau} =g_{\tau\sigma}$,are $g_{\tau\tau},g_{\tau\varphi} =g_{\varphi\tau}$, so finally :
$\mathcal{L}_\xi g_{\tau\tau}=\xi^r\partial_r g_{\tau\tau}+ 2(g_{\tau\tau}\partial_\tau\xi^\tau+g_{\tau\varphi}\partial_\tau\xi^\varphi )\tag{2}$
The first term is equal to :
$$\xi^r\partial_r g_{\tau\tau} = [2 GJ \Omega^2][-2\sqrt{1+r^2}\cos\tau][2(\Lambda^2-1)r]\tag{3}$$
The second term is equal to :
$$2g_{\tau\tau}\partial_\tau\xi^\tau = [2 GJ \Omega^2]2[-1 +(\Lambda^2-1)r^2][\frac{2r\cos\tau}{\sqrt{1+r^2}}]\tag{4}$$
The third term is equal to :
$$2g_{\tau\varphi}\partial_\tau\xi^\varphi = [2 GJ \Omega^2]2[\Lambda^2r][\frac{2\cos\tau}{\sqrt{1+r^2}}]\tag{5}$$
Finally we have :
$$\mathcal{L}_\xi g_{\tau\tau}= \\ \frac{(8 GJ \Omega^2)(r \cos\tau)}{\sqrt{1+r^2}}[-(1+r^2)(\Lambda^2-1)+(-1 +(\Lambda^2-1)r^2)+ \Lambda^2]=0\tag{6}$$
Best Answer
Got it for the 2 sphere.
$\xi=\xi^{\theta}\partial_{\theta}+\xi^{\phi}\partial_{\phi}=-A L_x + B L_y + C L_z$
A, B, C are the same integration constants as in the question and $L_x, L_y, L_z$ are the angular momentum operators written in the spherical polar coordinates.