We write this out in detail, to make sure that all the definitions work out in exactly the correct manner. Recall that $[X/G]$ is the quotient stack associated to the groupoid in algebraic spaces $(X, G \times_B X, s, t, c)$, see Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-quotient-stack}. This means that $[X/G]$ is the stackification of the category fibred in groupoids $[X/_{\!p}G]$ associated to the functor $$ (\Sch/S)_{fppf} \longrightarrow \textit{Groupoids}, \quad U \longmapsto (X(U), G(U) \times_{B(U)} X(U), s, t, c) $$ where $s(g, x) = x$, $t(g, x) = a(g, x)$, and $c((g, x), (g', x')) = (m(g, g'), x')$. By the construction of Categories, Example \ref{categories-example-functor-groupoids} an object of $[X/_{\!p}G]$ is a pair $(U, x)$ with $x \in X(U)$ and a morphism $(f, g) : (U, x) \to (U', x')$ of $[X/_{\!p}G]$ is given by a morphism of schemes $f : U \to U'$ and an element $g \in G(U)$ such that $a(g, x) = x' \circ f$. Hence we can define a $1$-morphism of stacks in groupoids $$ F_p : [X/_{\!p}G] \longrightarrow [[X/G]] $$ by the following rules: On objects we set $$ F_p(U, x) = (U, \pi \circ x, G \times_{B, \pi \circ x} U, a \circ (\text{id}_G \times x)) $$ This makes sense because the diagram $$ \xymatrix{ G \times_{B, \pi \circ x} U \ar[d] \ar[r]_{\text{id}_G \times x} & G \times_{B, \pi} X \ar[r]_-a & X \ar[d]^\pi \\ U \ar[rr]^{\pi \circ x} & & B } $$ commutes, and the two horizontal arrows are $G$-equivariant if we think of the fibre products as trivial $G$-torsors over $U$, resp.\ $X$. On morphisms $(f, g) : (U, x) \to (U', x')$ we set $F_p(f, g) = (f, R_g)$ where $R_g$ denotes right translation by $g$. More precisely, the morphism of $F_p(f, g) : F_p(U, x) \to F_p(U', x')$ is given by the cartesian diagram $$ \xymatrix{ G \times_{B, \pi \circ x} U \ar[d] \ar[r]_{R_{g^{-1}}} & G \times_{B, \pi \circ x'} U' \ar[d] \\ U \ar[r]^f & U' } $$ where $R_{g^{-1}}$ on $T$-valued points is given by $$ R_{g^{-1}}(g', u) = (m(g', i(g)), f(u)) $$ To see that this works we have to verify that $$ a \circ (\text{id}_G \times x) = a \circ (\text{id}_G \times x') \circ R_{g^{-1}} $$ which is true because the right hand side applied to the $T$-valued point $(g', u)$ gives the desired equality \begin{align*} a((\text{id}_G \times x')(m(g', i(g)), f(u))) & = a(m(g', i(g)), x'(f(u))) \\ & = a(g', a(i(g), x'(f(u)))) \\ & = a(g', x(u)) \end{align*} because $a(g, x) = x' \circ f$ and hence $a(i(g), x' \circ f) = x$. \medskip\noindent By the universal property of stackification from Stacks, Lemma \ref{stacks-lemma-stackify-groupoids-universal-property} we obtain a canonical extension $F : [X/G] \to [[X/G]]$ of the $1$-morphism $F_p$ above. We first prove that $F$ is fully faithful. To do this, since both source and target are stacks in groupoids, it suffices to prove that the $\mathit{Isom}$-sheaves are identified under $F$. Pick a scheme $U$ and objects $\xi, \xi'$ of $[X/G]$ over $U$. We want to show that $$ F : \mathit{Isom}_{[X/G]}(\xi, \xi') \longrightarrow \mathit{Isom}_{[[X/G]]}(F(\xi), F(\xi')) $$ is an isomorphism of sheaves. To do this it suffices to work locally on $U$, and hence we may assume that $\xi, \xi'$ come from objects $(U, x)$, $(U, x')$ of $[X/_{\!p}G]$ over $U$; this follows directly from the construction of the stackification, and it is also worked out in detail in Groupoids in Spaces, Section \ref{spaces-groupoids-section-explicit-quotient-stacks}. Either by directly using the description of morphisms in $[X/_{\!p}G]$ above, or using Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-morphisms} we see that in this case $$ \mathit{Isom}_{[X/G]}(\xi, \xi') = U \times_{(x, x'), X \times_S X, (s, t)} (G \times_B X) $$ A $T$-valued point of this fibre product corresponds to a pair $(u, g)$ with $u \in U(T)$, and $g \in G(T)$ such that $a(g, x \circ u) = x' \circ u$. (Note that this implies $\pi \circ x \circ u = \pi \circ x' \circ u$.) On the other hand, a $T$-valued point of $\mathit{Isom}_{[[X/G]]}(F(\xi), F(\xi'))$ by definition corresponds to a morphism $u : T \to U$ such that $\pi \circ x \circ u = \pi \circ x' \circ u : T \to B$ and an isomorphism $$ R : G \times_{B, \pi \circ x \circ u} T \longrightarrow G \times_{B, \pi \circ x' \circ u} T $$ of trivial $G_T$-torsors compatible with the given maps to $X$. Since the torsors are trivial we see that $R = R_{g^{-1}}$ (right multiplication) by some $g \in G(T)$. Compatibility with the maps $a \circ (1_G, x \circ u), a \circ (1_G, x' \circ u) : G \times_B T \to X$ is equivalent to the condition that $a(g, x \circ u) = x' \circ u$. Hence we obtain the desired equality of $\mathit{Isom}$-sheaves. \medskip\noindent Now that we know that $F$ is fully faithful we see that Stacks, Lemma \ref{stacks-lemma-characterize-essentially-surjective-when-ff} applies. Thus to show that $F$ is an equivalence it suffices to show that objects of $[[X/G]]$ are fppf locally in the essential image of $F$. This is clear as fppf torsors are fppf locally trivial, and hence we win.