The proof of existence of the Levi-Civita connection for pseudo-Riemannian manifolds uses heavily the fact that the metric is non-degenerate - so that $\nabla_XY$ is characterized by all the values $\langle \nabla_XY,Z\rangle$ via the Koszul formula.
I tried to come up with an example of a manifold with a degenerate metric which doesn't have the Levi-Civita connection, something like the flat metric $$\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$on $\Bbb R^2$, but I'm not sure how to come up with a contradiction (there should be an easier way than brute forcing my way through the Koszul formula). Is this the right way? Thanks.
If $g$ is a degenerate metric, then typically there will be lots of connections that are symmetric and compatible with $g$. For example, let $g$ be the constant-coefficient degenerate metric on $\mathbb R^2$ that you mentioned. Then a straightforward computation shows that the standard Euclidean connection $\nabla$ is compatible with $g$. Since it's symmetric (which has nothing to do with the metric), it would qualify as a "Levi-Civita connection" for $g$.
Now define a new connection $\widetilde\nabla$ by $$ \widetilde\nabla_X Y = \nabla_X Y + \langle X,Y\rangle \frac{\partial}{\partial y}. $$ Then $\widetilde\nabla$ is still symmetric and compatible with $g$ (because the vector field $\partial/\partial y$ is orthogonal to everything with respect to $g$).