The simple answer is: 

**You can't project to a null hypersurface**

This is tied to the fact that the "Lorentzian normal" vector field is in fact a tangent vector field. 

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In more details: 

Start with a tangent vector field $T^\nu$ to the image of $\Sigma$ in your Lorentzian manifold. If you lower the index by the metric $g_{\mu\nu}$, what you get is no longer necessarily a "tangent one form". Therefore it no longer makes sense to "pull it back". 

An explicit example is taking the null foliation of $\mathbb{R}^{1,1}$. Let $u$, $v$ be functions such that the Minkowski metric looks like
$$ 2 \mathrm{d}u ~\mathrm{d}v $$
And consider the vector field $\partial_u$ which is null, and is tangent to the level sets of $v$. Its metric dual is $2 \mathrm{d}v$ which restricts to zero (as a one form) on the level sets. 

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You should think that the induced connection in the non-degenerate setting really is saying the following:

The metric gives a canonical (unique) splitting of $TM$ along $\Sigma$ into $T \Sigma$ and $N\Sigma$, the normal bundle. Then the connection on $M$ induces a connection on $\Sigma$ by keeping only the part that is in $T \Sigma$ and tossing out everything that is in $N\Sigma$.

In the case $\Sigma$ is null, there is no unique splitting! In fact, $N\Sigma$ by the metric definition is part of $T\Sigma$! What you are missing is indeed just the projection: if you fix any choice of a transversal distribution to $T\Sigma$ in $TM$, then you are done: by linear algebra you can decompose any vectorfield in $TM$ into a portion in $T\Sigma$ and a portion in the transversal distributions. In the nondegenerate case the transversal distribution is provided by the normal bundle. In the degenerate case you don't have anything. (Which is why computations always fix, first and foremost, a null frame: once you fix a frame then you have a preferred transversal vector field and the theory goes through. But the choice of null frame is non-unique, which is why you have the entire theory of the GHP calculus.)