Let us write $\mathcal{U} = (\Sch/U)_{fppf}$. By our definition of algebraic stacks the $1$-morphism $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces. Hence in particular the 2-fibre product $$ \mathcal{U}_{x_0} = (\Sch/\Spec(k))_{fppf} \times_\mathcal{X} \mathcal{U} $$ is representable by an algebraic space $U_{x_0}$ over $\Spec(k)$. Then $U_{x_0} \to \Spec(k)$ is smooth and surjective (in particular $U_{x_0}$ is nonempty). By Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-separable-closed-points-dense} we can find a finite extension $l \supset k$ and a point $\Spec(l) \to U_{x_0}$ over $k$. We have $$ (\mathcal{F}_{\mathcal{X}, k , x_0})_{l/k} = \mathcal{F}_{\mathcal{X}, l, x_{l, 0}} $$ by Lemma \ref{lemma-change-of-field} and the fact that $\mathcal{X}$ satisfies (RS). Thus we see that $$ T\mathcal{F} \otimes_k l \cong T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}} \quad\text{and}\quad \text{Inf}(\mathcal{F}) \otimes_k l \cong \text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}) $$ by Formal Deformation Theory, Lemmas \ref{formal-defos-lemma-tangent-space-change-of-field} and \ref{formal-defos-lemma-inf-aut-change-of-field} (these are applicable by Lemmas \ref{lemma-algebraic-stack-RS} and \ref{lemma-deformation-category} and Remark \ref{remark-deformation-category-implies}). Hence it suffices to prove that $T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}})$ have finite dimension over $l$. Note that $x_{l, 0}$ comes from a point $u_0$ of $\mathcal{U}$ over $l$. \medskip\noindent We interrupt the flow of the argument to show that the lemma for infinitesimal automorphisms follows from the lemma for tangent spaces. Namely, let $\mathcal{R} = \mathcal{U} \times_\mathcal{X} \mathcal{U}$. Let $r_0$ be the $l$-valued point $(u_0, u_0, \text{id}_{x_0})$ of $\mathcal{R}$. Combining Lemma \ref{lemma-fibre-product-deformation-categories} and Formal Deformation Theory, Lemma \ref{formal-defos-lemma-deformation-functor-diagonal} we see that $$ \text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}) \subset T\mathcal{F}_{\mathcal{R}, l, r_0} $$ Note that $\mathcal{R}$ is an algebraic stack, see Algebraic Stacks, Lemma \ref{algebraic-lemma-2-fibre-product-general}. Also, $\mathcal{R}$ is representable by an algebraic space $R$ smooth over $U$ (via either projection, see Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}). Hence, choose an scheme $U'$ and a surjective \'etale morphism $U' \to R$ we see that $U'$ is smooth over $U$, hence locally of finite type over $S$. As $(\Sch/U')_{fppf} \to \mathcal{R}$ is surjective and smooth, we have reduced the question to the case of tangent spaces. \medskip\noindent The functor (\ref{equation-functoriality}) $$ \mathcal{F}_{\mathcal{U}, l, u_0} \longrightarrow \mathcal{F}_{\mathcal{X}, l, x_{l, 0}} $$ is smooth by Lemma \ref{lemma-formally-smooth-on-deformation-categories}. The induced map on tangent spaces $$ T\mathcal{F}_{\mathcal{U}, l, u_0} \longrightarrow T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}} $$ is $l$-linear (by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-k-linear-differential}) and surjective (as smooth maps of predeformation categories induce surjective maps on tangent spaces by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-smooth-morphism-essentially-surjective}). Hence it suffices to prove that the tangent space of the deformation space associated to the representable algebraic stack $\mathcal{U}$ at the point $u_0$ is finite dimensional. Let $\Spec(R) \subset U$ be an affine open such that $u_0 : \Spec(l) \to U$ factors through $\Spec(R)$ and such that $\Spec(R) \to S$ factors through $\Spec(\Lambda) \subset S$. Let $\mathfrak m_R \subset R$ be the kernel of the $\Lambda$-algebra map $\varphi_0 : R \to l$ corresponding to $u_0$. Note that $R$, being of finite type over the Noetherian ring $\Lambda$, is a Noetherian ring. Hence $\mathfrak m_R = (f_1, \ldots, f_n)$ is a finitely generated ideal. We have $$ T\mathcal{F}_{\mathcal{U}, l, u_0} = \{\varphi : R \to l[\epsilon] \mid \varphi \text{ is a } \Lambda\text{-algebra map and } \varphi \bmod \epsilon = \varphi_0\} $$ An element of the right hand side is determined by its values on $f_1, \ldots, f_n$ hence the dimension is at most $n$ and we win. Some details omitted.