This is a purely mathematical question (I am starting to study Minkowski spaces and I don't know how to prove that there are only 2 time-orientations).
Let $V$ be a vector space and $(,):V\times V\to\mathbb{R}$ be a non-degenerated, bilinear, symmetric form. Suppose $T=\{v\in V:(v,v)<0\}$ is non-empty (therefore $(,)$ is not an inner product but a pseudo inner product). (This is the set of time-like vectors).
On $T$ the following equivalent relation is defined: for $u,v\in T$, $u\sim v$ iff $(u,v)<0$.
Prove that there are only two equivalence classes, i.e. show that if $u,v,w\in T$, such that $(u,v)>0$ and $(u,w)>0$, then $(v,w)<0$.
If it has already been shown that $\sim$ is an equivalence relation (as it is part of the question, I assume it is true), and if we introduce the additional condition $(u,v)\neq 0\;\forall\, u,v\in T,$ then the statement can be proven as follows:
Take an arbitrary element $u_1$ of $T.$
It can easily be seen that $-u_1\in T,$ because $(-u_1,-u_1)=(-1)(-1)(u_1,u_1)=(u_1,u_1),$ due to the bilinearity of $(\, ,\,).$
The equivalence class of all elements in $T$, which are equivalent to $u_1,$ is $T_1$. If $v\in T$ is not an element of $T_1,$ then $(u_1,v)>0.$ We explicitly excluded the possibility $(u_1,v)=0.$ But $(u_1,v)>0$ means $(-u_1,v)<0.$ So each element of $T,$ which is not equivalent to $u_1,$ is necessarily equivalent to $-u_1,$ which means that we have exactly two equivalence classes, one with the representative $u_1$ and one with the representative $-u_1.$