Let consider the space $\mathcal{O}:=\{X\in \mathcal{M}_4(\mathbb{R}) \ / \ ^tX DX=D \}$ with $D=diag(1,1,1,-1)$.
Maybe we can apply the submersion's criterion.
We take $f : \mathcal{M}_4(\mathbb{R}) \to \mathcal{S}_4(\mathbb{R})\simeq \mathbb{R}^{10}, \ X\mapsto \ ^tX DX-D$ and we want to prove that $\mathrm{d}f_X$ is a surjective map.
Let $H \in \mathcal{M}_4(\mathbb{R})$ :
$f(X+H) = \ ^t(X+H)D(X+H) - D = f(X) + \ ^tXDH + \ ^tHDX + \text{o}(\Vert H \Vert)$.
So $\mathrm{d}f_X(H)= \ ^tXDH + \ ^tHDX$. Then I do not know how to continue.
Note that the dimension of $\mathcal{O}$ is $6$.
Thanks in advance !
Hint As Daniel Kruse points out in the comments, you do not need $f$ to be a submersion; it's sufficient to show that it is a submersion on $\mathcal O$, i.e., that $$df_X : T_X \mathcal M_4(\Bbb R) \to T_{f(X)} S_4(\Bbb R) \cong S_4(\Bbb R)$$ is surjective for all $X \in \mathcal O$. (Indeed, $df_0$ is the zero map and so is not surjective.)
So, we need to show that for any $X \in \mathcal O$ that any symmetric matrix $S$ can be written in the form $$S = {}^t\! XDH + {}^t\! HDX$$ for some $H$. This equation is complicated by the fact that $D$ is not the identity matrix (which is the case for the orthogonal group of the standard inner product). On the other hand, ${}^t D = D$ and $D^2 = I$, which suggests that we can simplify our condition by writing the unknown $H$ as $DA$ (i.e., setting $A = D^{-1} H = DH$).
Then, the surjectivity condition is that we can find an $A$ such that $$S = {}^t\! XD(DA) + {}^t (DA)DX,$$ and this expression simplifies.
Incidentally, $\mathcal{O}$ is a group under matrix multiplication; in fact it is the Lorentz group $O(1, 3)$ of linear transformations that fix a light cone in Minkowski space.