Let $[[(a,b)]]_{\ominus}$, $[[(c,d)]]_{\ominus}$ and $[[(e,f)]]_{\ominus}$ be formally constructed rational numbers - equivalence classes of ordered pairs under the $(z_1,z_2)\,{\ominus}\,(z_3,z_4){\iff}z_1*z_4=z_2*z_3$ relation, where $z_1,z_3\,{\in}\,\mathbb{Z}$ and $z_2,z_4\,{\in}\,\mathbb{Z}/{\{0\}}$. For convenience, I'll use $[a,b]$, $[c,d]$ and $[e,f]$ to mean those equivalence classes.
Let $(\mathbb{Q},\mathbb{Q},{\le})$ be a relation defined by $[m_1,n_1]\,{\le}\,[m_2,n_2]{\iff}(n_1n_2<0\,{\land}\,m_1n_2\,{\le}_{\mathbb{Z}}\,m_2n_1)\,{\lor}\,(n_1n_2>0\,{\land}\,m_2n_1\,{\le}_{\mathbb{Z}}\,m_1n_2)$.
How to prove that $\Big(\big(([a,b]\,{\le}\,[c,d])\,{\land}\,([c,d]\,{\le}\,[e,f])\big){\implies}([a,b]\,{\le}\,[e,f])\Big)$?
Actually if you restrict denominators to positive integers, you don't have to treat cases. If $\frac{a}{b}\leq\frac{c}{d}$ and $\frac{c}{d}\leq\frac{e}{f}$ then $ad\leq cb$ and $cf\leq ed$ so $adf\leq cbf\leq edb$ and $adf\leq edb$. Simplifying, we get $af\leq eb$, i.e. $\frac{a}{b}\leq\frac{e}{f}$.