This is parallel transport. Note that you only get an explicit formula for the isomorphism between $F_1^*H(\mathbb{T})$ and $F_2^*H(\mathbb{T})$. To get explicit formulas over other thickenings of $\mathbb{T}$, you would need lifts of both $\mathbb{T}$ and also of the Frobenius lifts. For convenience, set $H_1=H(\mathbb{T})$. We will assume that $\mathbb{T}$ has \'etale co-ordinates $x_1,\ldots,x_n$.
Since $H_{\mathbb{T}}$ is a crystal over $\mathbb{T}$, there is a flat connection $\nabla:H_1\to H_1\otimes\Omega^1_{\mathbb{T}/W_2(k)}$. We now use Taylor expansions! Note that, for any $a\in\mathcal{O}_{\mathbb{T}}$, $\delta(a)=F_1(a)-F_2(a)$ is in $p\mathcal{O}_{\mathbb{T}}$; in particular, $\delta(a)^2=0$. Now, for any $h\in H_1$, the isomorphism sends $F_1^*h$ to $F_2^*h+\sum_{i=1}^n\nabla(\partial_i)(m)\delta(x_i)$. This is just a Taylor expansion up to the first degree.
This is parallel transport. Note that you only get an explicit formula for the isomorphism between $F_1^*H(\mathbb{T})$ and $F_2^*H(\mathbb{T})$. To get explicit formulas over other thickenings of $\mathbb{T}$, you would need lifts of both $\mathbb{T}$ and also of the Frobenius lifts. For convenience, set $H_1=H(\mathbb{T})$. We will assume that $\mathbb{T}$ has \'etale co-ordinates $x_1,\ldots,x_n$.
Since $H_{\mathbb{T}}$ is a crystal over $\mathbb{T}$, there is a flat connection $\nabla:H_1\to H_1\otimes\Omega^1_{\mathbb{T}/W_2(k)}$. We now use Taylor expansions! Note that, for any $a\in\mathcal{O}_{\mathbb{T}}$, $\delta(a)=F_1(a)-F_2(a)$ is in $p\mathcal{O}_{\mathbb{T}}$; in particular, $\delta(a)^2=0$. Now, for any $h\in H_1$, the isomorphism sends $F_1^*h$ to $F_2^*h+\sum_{i=1}^n\nabla(\partial_i)(m)\delta(x_i)$.