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). 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). To me, this seems to almost have to be the start of some sort of long exact sequence. 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 homotopical algebra? More generally, is there a more conceptual way of viewing the above sequence?