You just compute! Wald is just skipping these steps because they’re “elementary” to work out, and also a good exercise.
First, take the exterior derivative of both sides of equation (3) (use the product rule on the LHS, and simplify… some cancellation will happen) to get
\begin{align}
\frac{re^{r/2M}}{4M^2}dr&=2XdX-2TdT.\tag{@}
\end{align}
So, if we plug in $X=T=0$ then the RHS vanishes, while on the LHS, we note that (3) implies $r=2M$. This is a non-zero coefficient of $\frac{e}{2M}$ on the left, and thus $(dr)|_{X=T=0}=0$.
Next, (4) gives you an equation for $dt$ in terms of $dX$ and $dT$. So, combined with the previous equation for $dr$ in terms of $dX,dT$, you ave two equations with two “unknowns”, and you can invert the system of equations to rewrite $dT,dX$ in terms of $dt,dr$. Once you do this calculation, you should be able to prove that in terms of the $(T,X,\theta,\phi)$ coordinate system, the static Killing field $\xi$ is given by
\begin{align}
\xi&=dT(\xi)\frac{\partial}{\partial T}+dX(\xi)\frac{\partial}{\partial X}\\
&=\frac{1}{4M}\left(X\frac{\partial}{\partial T}+ T\frac{\partial}{\partial X}\right).\tag{$*$}
\end{align}
Ok very strictly speaking, we only know that this equality holds in region I of the spacetime, which is the common domain of definition of the two coordinate systems $(T,X,\theta,\phi),(t,r,\theta,\phi)$. But now, we observe that the vector field on the right of $(*)$ is well-defined and actually analytic with respect to the global (up to the usual caveats with spherical coordinates) $(T,X,\theta,\phi)$ coordinates. So, we shall use $(*)$ as our definition of the vector field $\xi$ on the maximally extended spacetime. Since the metric $g$ is also analytic, and we know for sure that $\xi$ is a Killing vector field in region I, it follows by uniqueness of analytic continuation that $\xi$ is a Killing field on the entire maximally extended spacetime (which vanishes if and only if $X=T=0$). I’m sure you could try to argue directly that the RHS of $(*)$ is indeed Killing everywhere, but using this analysis fact is much quicker.
Next, to prove collinearity, we need to use the metric-induced isomorphism. We have
\begin{align}
\begin{cases}
g^{\flat}\left(\frac{\partial}{\partial T}\right)&=g_{TT}\,dT+g_{TX}\,dX+g_{T\theta}\,d\theta
+g_{T\phi}\,d\phi=-\alpha(r)\,dT\\
g^{\flat}\left(\frac{\partial}{\partial X}\right)&=g_{XT}\,dT+g_{XX}\,dX+g_{X\theta}\,d\theta+g_{X\phi}\,d\phi=\alpha(r)\,dX,
\end{cases}
\end{align}
where $\alpha(r):=\frac{32M^3e^{-r/2M}}{r}$. Therefore, applying $g^{\flat}$ to $(*)$ above (i.e calculating $\xi_a$), we get
\begin{align}
g^{\flat}(\xi)&=\frac{\alpha(r)}{4M}\left(-X\,dT+T\,dX\right)\equiv A(r)\left(-X\,dT+T\,dX\right).
\end{align}
Compare this with (@) above, which says $dr=B(r)\left(X\,dX-T\,dT\right)$, for some positive function $B(r)$. Hence, we see that on the null lines $X=\pm T$ (which corresponds to $r=2M$), we have $g^{\flat}(\xi)=\pm\frac{A(2M)}{B(2M)}\,dr$.
