Say $(U, T, \delta)$ is an object of $\text{Cris}(X/S)$. Let $(U, T', \delta')$ be the infinitesimal thickening of $T$ by $(\Omega_{X/S})_T = \Omega_{T/S, \delta}$ constructed in Remark \ref{remark-first-order-thickening}. It comes with projections $p_0, p_1 : T' \to T$ and a diagonal $i : T \to T'$. By assumption we get isomorphisms $$ p_0^*\mathcal{F}_T \xrightarrow{c_0} \mathcal{F}_{T'} \xleftarrow{c_1} p_1^*\mathcal{F}_T $$ of $\mathcal{O}_{T'}$-modules. Pulling $c = c_1^{-1} \circ c_0$ back to $T$ by $i$ we obtain the identity map of $\mathcal{F}_T$. Hence if $s \in \Gamma(T, \mathcal{F}_T)$ then $\nabla(s) = p_1^*s - c(p_0^*s)$ is a section of $p_1^*\mathcal{F}_T$ which vanishes on pulling back by $i$. Hence $\nabla(s)$ is a section of $$ \mathcal{F}_T \otimes_{\mathcal{O}_T} \Omega_{T/S, \delta} $$ because this is the kernel of $p_1^*\mathcal{F}_T \to \mathcal{F}_T$ as $\mathcal{O}_{T'} = \mathcal{O}_T \oplus \Omega_{T/S, \delta}$ by construction. It is easily verified that $\nabla(fs) = f\nabla(s) + s \otimes \text{d}(f)$ using the description of $\text{d}$ in Remark \ref{remark-first-order-thickening}. \medskip\noindent The collection of maps $$ \nabla : \Gamma(T, \mathcal{F}_T) \to \Gamma(T, \mathcal{F}_T \otimes_{\mathcal{O}_T} \Omega_{T/S, \delta}) $$ so obtained is functorial in $T$ because the construction of $T'$ is functorial in $T$. Hence we obtain a connection. \medskip\noindent To show that the connection is integrable we consider the object $(U, T'', \delta'')$ constructed in Remark \ref{remark-second-order-thickening}. Because $\mathcal{F}$ is a sheaf we see that $$ \xymatrix{ q_0^*\mathcal{F}_T \ar[rr]_{q_{01}^*c} \ar[rd]_{q_{02}^*c} & & q_1^*\mathcal{F}_T \ar[ld]^{q_{12}^*c} \\ & q_2^*\mathcal{F}_T } $$ is a commutative diagram of $\mathcal{O}_{T''}$-modules. For $s \in \Gamma(T, \mathcal{F}_T)$ we have $c(p_0^*s) = p_1^*s - \nabla(s)$. Write $\nabla(s) = \sum p_1^*s_i \cdot \omega_i$ where $s_i$ is a local section of $\mathcal{F}_T$ and $\omega_i$ is a local section of $\Omega_{T/S, \delta}$. We think of $\omega_i$ as a local section of the structure sheaf of $\mathcal{O}_{T'}$ and hence we write product instead of tensor product. On the one hand \begin{align*} q_{12}^*c \circ q_{01}^*c(q_0^*s) & =  q_{12}^*c(q_1^*s - \sum q_1^*s_i \cdot q_{01}^*\omega_i) \\ & = q_2^*s - \sum q_2^*s_i \cdot q_{12}^*\omega_i - \sum q_2^*s_i \cdot q_{01}^*\omega_i + \sum q_{12}^*\nabla(s_i) \cdot q_{01}^*\omega_i \end{align*} and on the other hand $$ q_{02}^*c(q_0^*s) = q_2^*s - \sum q_2^*s_i \cdot q_{02}^*\omega_i. $$ From the formulae of Remark \ref{remark-second-order-thickening} we see that $q_{01}^*\omega_i + q_{12}^*\omega_i - q_{02}^*\omega_i = \text{d}\omega_i$. Hence the difference of the two expressions above is $$ \sum q_2^*s_i \cdot \text{d}\omega_i - \sum q_{12}^*\nabla(s_i) \cdot q_{01}^*\omega_i $$ Note that $q_{12}^*\omega \cdot q_{01}^*\omega' = \omega' \wedge \omega = - \omega \wedge \omega'$ by the definition of the multiplication on $\mathcal{O}_{T''}$. Thus the expression above is $\nabla^2(s)$ viewed as a section of the subsheaf $\mathcal{F}_T \otimes \Omega^2_{T/S, \delta}$ of $q_2^*\mathcal{F}$. Hence we get the integrability condition.