Let $X$ be a finite-type scheme over a field $k$. Let $G$ be a finite-type group scheme over $k$; we write $G_X$ for the base-change of $G$ from $\operatorname{Spec}(k)$ to $X$.

Suppose $f : Y \rightarrow X$ is a $G_X$-torsor for the fppf topology (i.e. there exists an $X$-group scheme action of $G_X$ on $Y$ such that the morphism $G_X \times_X Y \rightarrow Y \times_X Y$ given on points by $(g,y) \mapsto (y,gy)$ is an isomorphism). Such a $Y$ gives a class in the fppf cohomology set $H^1(X,G_X)$ that classifies fppf $G_X$-torsor sheaves over $X$ (this is defined with Cech cohomology).

Consider the specialization map $$ s : X(k) \rightarrow H^1(k,G) $$ that sends $x : \operatorname{Spec}(k) \rightarrow X$ to the pull-back of $Y$ by $x$, which gives a $G$-torsor over $\operatorname{Spec}(k)$, hence an element of $H^1(k,X)$ (defined similarly as above). Note: if $k$ is perfect, the fppf cohomology set $H^1(k,G)$ may be identified with the Galois cohomology set $H^1(k,G(\overline{k}))$.

Fixing a $y \in Y(k)$ (if it exists) we obtain an exact sequence of pointed sets $$ 0 \rightarrow G(k) \rightarrow Y(k) \stackrel{f}{\rightarrow} X(k) \stackrel{s}{\rightarrow} H^1(k,G) $$ where $G(k) \rightarrow Y(k)$ is just the inclusion of the fiber above $y$ (if $y$ doesn't exist it is the unique map between empty sets). From the looks of it, I'd say that this has to be the start of a long exact sequence of some kind. The question only is: what kind? I don't see an obvious way of continuing it, since $Y$ doesn't necessarily carry any group structure so as to give meaning to the expression $H^1(k,Y)$.

Question: Is this exact sequence part of a long exact sequence? For instance, are we witnessing some instantiation of non-commutative cohomology or even homotopy theory? More generally, is there a more conceptual way of viewing the above sequence?