Interpretation of components of energy-momentum flux near a null surface Let $k^a$ be the normal to a null surface and $l^a$ be the auxiliary null vector satisfying $l^a k_a=-1$ (see, for instance, the textbook A Relativist's Toolkit by Poisson).
I wanted to understand the physical interpretation of the components $T_{ab}l^a l^b,T_{ab}l^a k^b$ and $T_{ab}k^a k^b$. For a non-null surface with unit normal $n^a$, $T^{ab}n_a n_b$ would be the flux through the surface of the component of momentum normal to the surface. Here $k^a$ is the normal. But it is also tangent to the surface. So I am not clear which component above would correspond to the flux through the surface of the component of momentum normal to the surface. Is there any reference where this issue has been discussed?
 A: This answer has been edited as I came to realise I had answered at first without understanding the question (the "no wait, I'm a moron!" experience). Therefore I will start with some unnecessary background.
Let's agree to call a co-vector $p_i$ perpendicular (and leave the term normal for metrically normal vectors; we shall soon see that they are the same) to a hypersurface $\Sigma$ defined locally by $f = 0$ for some smooth function $f$ if $p_i$ is colinear with $df$ on all of $\Sigma$. Similarly a vector $t^i$ will be referred to as tangential if for all $q \in \Sigma$ we have $t^i|_q \in T_q\Sigma$. Note that it is immediately obvious that any perpendicular co-vector must be normal to all tangential vectors. Thus for non-null hypersurfaces the (unit) perpendicular and the (unit) normal coincide and by continuity the perpendicular is an appropriate choice of normal also for null hypersurfaces. 
For null hypersurfaces there is a $t^i$ corresponding to each $p_i$ such that $t_i = p_i$, which is the issue at hand in your question: by the above reasoning it would seem that $k^i$ flux is both tangential and perpendicular/normal (having shown above that they are the same).

The directed surface element of any hypersurface, including a null one, can be written as (see e.g. my answer to this question)
$$
d\Sigma_i = \varepsilon_{i\vec{J}}\omega^{\vec{J}}|_{\Sigma},
$$
where $\varepsilon$ denotes the Levi-Civita tensor, $\vec{J}$ denotes strictly increasing multi-indices, $\omega^i$ is the local frame of $T^*M$, and $|_{\Sigma}$ denotes projection onto $\Sigma$, defined for $T^*M \to T^*\Sigma$ in local coordinates $(f,u,x,y)$ where $u$, $x$, and $y$ are coordinates on $\Sigma$, simply by $df = 0$. In the corresponding coordinate frame $k^f \equiv 0$ since $k^i$ is tangential to $\Sigma$, whence it is easy to see that $k^id\Sigma_i \equiv 0$. This tells us that the flux defined by $k^i$ does not cross $\Sigma$.
Since $k_i$ is colinear with $df$ we have $k_i = \varrho df = \varrho\delta^f_i$ whence $\ell^i$ contains the term $-\varrho^{-1}\partial_f = -\varrho^{-1}\delta^i_f$.
Thus
$$
\ell^id\Sigma_i = -\varrho^{-1}\varepsilon_{f\vec{J}}\omega^{\vec{J}}|_{\Sigma} \neq 0.
$$
Thus the flux defined by $\ell^i$ obviously does cross $\Sigma$, which should come as no surprise since $\ell^i$ is not tangent to $\Sigma$. But is there any reason to consider $\ell^i$ as "normal" to $\Sigma$? There is a piori no unique way of defining such a normal, but if we consider 3+1 formalism the foliation of spacetime induces a foliation of $\Sigma$, and (as you may know) $\ell^i$ can be defined uniquely (up to normalization with respect to $k^i$) as the null vector non-colinear with $k^i$ that is normal to the leaves.
For take a local Lorentz frame where $e_0$ is the normal to the 3+1 foliation, and $e_1,e_2,e_3$ spans the leaves. We can select $e_1$ such that $k^i = N(\delta^i_0 + \delta^i_1)$ (here $N$ is the lapse function, selected by demanding that $k^i = x^i{}_{,t}$, where $t$ parametrizes the foliation). Then $\Sigma$ is spanned by $k^i$, $\delta^i_2$, and $\delta^i_3$ and $\ell^i = \frac{1}{2N}(\delta^i_0 - \delta^i_1)$. For clarity I show below that the normalization by $N$ cannot be used to solve for $\varrho$, which should come as no surprise since $\varrho$ depends on $f$

It can be shown that $k^i$ must be a null geodesic velocity vector, whence we can choose $k^i = \partial_u$ without restriction. In the local Lorentz frame this gives us $\varrho df = k_i = N(\delta^0_i - \delta^1_i)$ and $-du = \ell_i = \frac{1}{2N}(\delta^0_i + \delta^1_0)$. We find
\begin{align*}
\omega^0 &= \frac{1}{2}\left(2N\ell_i + \frac{1}{N}k_i\right) &
\omega^1 &= \frac{1}{2}\left(2N\ell_i - \frac{1}{N}k_i\right) \\
&= -\frac{1}{2}\left(2Ndu - \frac{\varrho}{N}df\right), &
&= -\frac{1}{2}\left(2Ndu + \frac{\varrho}{N}df\right).
\end{align*}
Projection onto $\Sigma$ yields
$$
\omega^0|_\Sigma = \omega^1|_\Sigma = -Ndu,
$$
whence
$$
\ell^id\Sigma_i = -du\wedge\omega^2\wedge\omega^3.
$$
Finally note that in the frame $(df,du,\omega^2,\omega^3)$ the metric takes the form
$$
g_{ij} = \begin{bmatrix}
0 & \varrho & 0 & 0 \\
\varrho & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$
so that $\sqrt{|g|} = \varrho$, and $-\varrho^{-1}\varepsilon_{f\vec{J}}\omega^{\vec{J}} = -du\wedge\omega^2\wedge\omega^3$.

Other natural choices of normals (in that they are normal to the induced foliation) are $e_0$ (timelike) and $e_1$ (spacelike), but the null vector $\ell^i$ is often used because it simplifies the formulae for describing $\Sigma$.
I found this (Gourgoulhon and Jaramillo, Physics Reports 423 (2006) 159 – 294) paywalled (I think) paper to be a pretty good source on null geometry in 3+1 formalism, but I am sure there are others. More information on projection along the null vector $\ell^i$ can be found there (but note that they use the opposite notation, thus their $\ell$ is your $k$, and vice versa).
