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).
