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. 

-------------------

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. 

-------------------

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.)

--------------------

To add a few more details:

> If $\gamma$ is degenerate, then it is usually said that there is no unique induced connection on $Σ$, however the pullback of the covariant derivative on $M$ can be defined the same way as above, except all the latin indices will be covariant incides, because $\gamma^{ij}$ doesn't exist to raise indices.

That is strictly not true. What you have defined is the the pull-back covariant derivative of the pullback of the cotangent bundle on $M$. This does not give you a covariant derivative on the cotangent bundle on $\Sigma$, since there is an ambiguity involved in how to "pushforward" $T^*\Sigma$ to $T^*M$. 

Revisiting the example with the two dimensional Minkowski space above, the issue is that given a one form tangent to $\Sigma$ being a level set of $v$, you can write it as, in the coordinate $u$, some function $f(u) ~\mathrm{d}u$. But you do not know how to canonically associate to this one form a one form on $M$: for any function $g(u,v)$, the one form $f(u) ~\mathrm{d}u + g(u,v) ~\mathrm{d}v$ restricts to your given one form on $\Sigma$. So knowing how to differentiate on the pull back of the cotangent bundle doesn't help you do anything with objects in $T^*\Sigma$. 

The same mistake cropped up in 

> So if we are given a tensor on $Σ$, we can express it in terms of greek indices (covariant latin indices can be made covariant greek indices the same way contravariant greeks can be made latin I detailed in the early part of my post)

Contravariant tensors on $\Sigma$ can be expressed in terms of "greek indices", but covariant tensors cannot, because of the ambiguity described above.