Given a group $G$ and a left $G$-set $X$, then we can make $X$ a right $G$-set defining the action as $xg:=g^{-1}x$ , or if you prefer we are considering the opposite group $G^{op}$ to make the left action a right action. Note that in this way $X$ is a $(G,G^{op})$-set, i.e. it is a left $G$-set and a right $G^{op}$-set.
Now, talking about $G$-torsors, there is the notion of contracted product, i.e. a product of torsors. In the literature the definition of the product is:
Given X a $(G,H)$-bitorsor and $Y$ a $(H,G)$-bitorsor then $X \times^H Y=\frac{X \times Y}{\Delta(H)}$ is a $G$-bitorsor
Here $\Delta(H)$ is the diagonal action, or equivalently the equivalence relation is defined by $(xh,y) \sim (x,hy)$. One can show that this product is associative.
Now, my question is, here there is a definition based on bitorsors, but what if, given $X,Y$ left $G$-torsor, i take the quotient of the fibre product of $X,Y$ by the equivalence relation $(gx,y) \sim (x,gy)$? Why this is not a left $G$-torsor?
In the case where $G$ is abelian, we can note that every $G$-torsor is a $G$-bitorsor and the product will be a $G$-bitorsor as well. But in the non abelian case? I suppose that there could be a problem of associativity with this product, but I don't know if it is the case and why.