I think you are slightly confused about the overall picture. Let's review it again:

All is about the construction of a suitable frames. By "frame", I mean a set of four basis vectors at each point of spacetime. Given a coordinate system over the spacetime, we already have one: the "coordinate basis" given by partial derivatives $\partial_\mu$ along each coordinate (e.g. $\{\partial_t,\partial_r,\partial_\theta,\partial_\phi \}$). However, the problem with this frame is that it is not orthonormal in a curved spacetime, since $\partial_\mu\cdot \partial_\nu=g_{\mu\nu}$. For various reasons, we prefer to work with orthonormal frames. Therefore we transform from the coordinate basis $\partial_\mu$ to another set of four basis vectors $e_{(a)}=e_{(a)}{}^\mu\partial_\mu$ for $a=\{0,1,2,3\}$ such that 
\begin{align}
e_{(a)}\cdot e_{(b)}=\eta_{(a)(b)}
\end{align}
I put parantheses around "$a,b$" indices to stress that they are "label indices" to be distinguished from coordinate indices $\mu$. 
For a Rimannian geometry it is natural to take $\eta_{(a)(b)}=\text{diag}(1,1,1,1)$, which implies that the four vectors are orthonormal, i.e. of unit norm and orthogonal to each other. For a Lorentzian spacetime, we instead typically consider $\eta_{(a)(b)}=\text{diag}(-1,1,1,1)$ so that $e_{(0)}$ is timelike while the other three are spacelike. 

However, in some situations (including black hole perturbation theory), we prefer to work with null basis vectors, i.e. $|e_{(0)}|^2=e_{(0)}\cdot e_{(0)}=0$ and similarly for the other three basis vectors. To form a complete basis, then one typically takes $e_{(0)}\cdot e_{(1)}=-1, e_{(2)}\cdot e_{(3)}=1$ and the rest of the inner products to be zero. Therefore the matrix $\eta_{(a)(b)}$ looks like what you wrote above [except that you have replaced (0,1) with (2,3)]. In Boyer-Lindquist coordinates for Kerr spacetime, a choice for the basis vectors is 
\begin{aligned}
l^{\mu} &=\frac{1}{\Delta}\left(r^{2}+a^{2}, \Delta, 0, a\right), \\
q^{\mu} &=\frac{1}{2 \rho^{2}}\left(r^{2}+a^{2},-\Delta, 0, a\right), \\
m^{\mu} &=\frac{1}{\sqrt{2}} \frac{1}{r+i a \cos \theta}\left(i a \sin \theta, 0,1, \frac{i}{\sin \theta}\right),
\end{aligned}
and $e_{(a)}{}^\mu$ is a 4$\times$4 matrix whose columns are the basis vectors, i.e. $e_{(a)}{}^\mu=\Big(l^\mu\,|\,n^\mu\,|\,m^{\mu}\,|\,\overline{m}^{\mu}\Big)$.


The inverse of $e_{(a)}{}^\mu$ is instead a set of four one-forms $e^{(a)}=e^{(a)}{}_\mu dx^\mu$ whose components are given by 
$$e^{(a)}{}_\mu=\eta^{(a)(b)}g_{\mu\nu}e_{(b)}{}^\nu$$

From this we see that as a column matrix  $e^{(a)}{}_\mu=\Big(-n_\mu\,|\,-l_\mu\,|\,\overline{m}^{\mu}\,|\,{m}^{\mu}\Big)$

Now it is very clear that $e^{(a)}{}_\mu l^\mu=(-n_\mu l^\mu, -l_\mu l^\mu,\overline{m}_\mu l^\mu,m_\mu l^\mu,)=(1,0,0,0)$. So as you see $l^{(a)}=e^{(a)}{}_\mu l^\mu$ the vector $l$ is the zeroth basis vector and hence contains no particularly useful information. Also obviously $l^{(a)}l_{(a)}=\eta_{(a)(b)}l^{(a)}l^{(b)}=\eta_{(0)(0)}=0$. this is another way to say that $l^\mu$ is by construction a null vector.

I hope this helps.