Consider the following group: $$\text{GL}^+(2,\mathbb{Q})=\{M\in\text{GL}(2,\mathbb{Q}):\det M>0\}$$ I'm trying to prove the following fact: $$\text{GL}^+(2,\mathbb{Q})=\bigsqcup_{\substack{x,z\in\mathbb{Q}\ \cap\ (0,+\infty)\\y\in\mathbb{Q}/\mathbb{Z}}}\text{SL}(2,\mathbb{Z})\begin{pmatrix}x&y\\0&z\end{pmatrix}$$
My only problem is to prove that the union in the RHS is a disjoint union. This is my attempt: given $x_1,x_2,z_1,z_2\in\mathbb{Q}\cap(0,+\infty)$ and $y_1,y_2\in\mathbb{Q}/\mathbb{Z},$ suppose that $$\text{SL}(2,\mathbb{Z})\begin{pmatrix} x_1&y_1\\ 0&z_1 \end{pmatrix}=\text{SL}(2,\mathbb{Z})\begin{pmatrix} x_2&y_2\\ 0&z_2 \end{pmatrix}$$ Considering the determinant, I immediately get $x_1z_1=x_2z_2.$ I can rewrite the preceding relation in the following way: $$ \frac{1}{x_1z_1}\text{SL}(2,\mathbb{Z})\begin{pmatrix} x_1&y_1\\ 0&z_1 \end{pmatrix}\begin{pmatrix} z_2&-y_2\\ 0&x_2 \end{pmatrix}=\text{SL}(2,\mathbb{Z})$$
$$\text{SL}(2,\mathbb{Z})\begin{pmatrix} \displaystyle\frac{z_2}{z_1}&\displaystyle-\frac{y_2}{z_1}+\frac{y_1}{z_2}\\[10pt] 0&\displaystyle\frac{x_2}{x_1} \end{pmatrix}=\text{SL}(2,\mathbb{Z})$$ $$\text{SL}(2,\mathbb{Z})\cdot M=\text{SL}(2,\mathbb{Z})$$
From this I should get that $M\in\text{SL}(2,\mathbb{Z})$ and in particular, using the hypothesis $x_1,x_2,z_1,z_2>0,$ that $z_1=z_2,x_1=x_2,$ hence $$M=\begin{pmatrix} \displaystyle 1 &\displaystyle-\frac{y_2+y_1}{z_1}\\[10pt] 0&\displaystyle1 \end{pmatrix}\in\Gamma(1)$$
Is it everything right? I don't know how to prove that $y_2-y_1$ is an integer.
There is a mistake in your question, it is $y\in \Bbb{Q/zZ}$ not $y\in \Bbb{Q/Z}$.
You got that $$\pmatrix{x_1&y_1\\0&z_1}=\pmatrix{1&n\\0&1} \pmatrix{x_2&y_2\\0&z_2}$$ so that $x_1=x_2,z_1=z_2$ and $y_1 = y_2+nz_2$ ie. $\pmatrix{x_1&y_1+z_1\Bbb{Z}\\0&z_1}=\pmatrix{x_2&y_2+z_2\Bbb{Z}\\0&z_2}$ as wanted.
It remains to prove that every $\pmatrix{u&v\\w&z}\in GL_2(\Bbb{Q})^+$ matrix is in one of the cosets. If $uw\ne 0$, write that $(u,w)=r (d,-c)$ for some coprime integers, take $a,b$ such that $ad-bc=1$ then look at $\pmatrix{a&b\\c&d}\pmatrix{u&v\\w&z}$. For the case $w=0$ there is nothing to do, for the case $u=0$ use $\pmatrix{0&1\\-1&0}$.