Recall the definition of allowed coefficients.
Definition
[OAI26], Definition 3.1 The basic coefficients are \[ \mathcal{L}_m^{\otimes t}\otimes_{A_m}\bigotimes_{i=1}^s D_m(\ell_i, b_i) \] for $s\geq 0$, $\ell_i\geq 0$ and $0\leq b_i\leq m\ell_i$, and $t\in \mathbb{Z}$.
The allowed coefficient is the smallest class of $A_m$-modules $\mathcal{C}_m$ containing the basic coefficients closed under finite direct sum, tensor product, taking equivariant direct summand, and finite equivariant filtrations whose successive quotients belong to the class
Lemma 3.1 (Weierstrass Preparation Theorem)
Let $R$ be a complete local ring, with maximal ideal $\mathfrak{m}$. Suppose $f$ is a formal power series over $R$, such that $f\not \equiv 0 \mod \mathfrak{m}$, then there is $v\in R[[t]]^\times$ and monic polynomial $W(t)=t^{s}+\sum_{i=0}^{s-1} b_i t^i$ for $b_i \in \mathfrak{m}$.
Lemma 3.2
The algebra $D_m(\ell, b)$ are free of rank $p^{n\ell-b}$ over $A_m$. Moreover, the construction of this subalgebra is independent of the choice of coordinate, i.e. suppose we have a change of coordinate $\alpha: A_m[[T]]\to A_m[[\alpha(T)]]$, the Frobenius and hence $D_m(\ell, b)$ are unchanged, with $D_m(\ell, b)/(x)=D_{m+1}(\ell, b)$. Lastly, every module in $\mathcal{C}_m$ are finite free over $A_m$, and modulo $x$ takes $\mathcal{C}_m$ to $\mathcal{C}_{m+1}$.
proof:
Since $[p^\ell]_F T=K(T^{p^{m\ell}})$ for some power series $K$, there is power series $f$ such that $f(T^{p^b})=[p^\ell]_F T$, and $f(t)=x^\ell t^{p^{m\ell-b}}+…$.
Note that $F$ is a deformation of $\Gamma_n$, hence modulo the maximal ideal recovers the $n$-series. In other word, $[p^\ell]_F T\equiv T^{n\ell}\mod \mathfrak{m}$. Therefore, $f(t)\equiv t^{p^{n\ell-b}}\mod \mathfrak{m}$.
By Weierstrass Preparation Theorem, $f(t)=v(t)W(t)$, with $W(t)=t^{p^{n\ell-b}}+…$. Since $W$ is monic, $A_m[t]/(W)$ is finite free over $A_m$, with rank $p^{n\ell-b}$.
Define map $A_m[t]/(W)\to A_m[[T]]/([p^\ell]_F T)$ via $t\mapsto T^{p^b}$, this is injective and has image $D_m(\ell, b)$, which shows the first statement.
For a coordinate change, we have $\alpha(T^{p^b})=\alpha(T)^{p^b}$, so the subalgebra is preserved.
We see that $[p^\ell]_F T$ over $A_m$ modulo $x$ is precisely $[p^\ell]_FT$ over $A_{m+1}$, so the subalgebra is preserved under restriction to higher height locus. Similarly, $\mathcal{L}_m/ (x)=\mathcal{L}_{m+1}$ as well.
Lastly, finite tensor product and finite direct sums preserves finite freeness. Since $A_m$ is local, by Kaplansky’s Theorem of Projective Modules, direct summand of finite free modules are finite free.
Now its left with the filtration condition. Given a finite filtration $F^t M\subseteq F^{t-1}\subseteq F^0M$, where $F^iM/F^{i+1}M\in \mathcal{C}_m$. There are exact sequences \[ 0\to F^{i+1}M\to F^iM\to F^iM/F^{i+1}M\to 0 \] The right term is by assumption finite free, hence $F^iM$ splits. Inducting on $i$, we are then left to show that $F^tM$ is finite free, which is the assumption.
Q.E.D.
We denote $F[p^\ell]:=\spec A_m[[T]]/([p^\ell]_F T)$ the scheme of $p^\ell$-torsions of $F$. There is inclusion $F[p^\ell]\to F[p^{\ell+1}]$, and we denote $F[p^\infty]$ the $p$-divisible group of $F$.
Now over, $B_m=A_m[1/x]$, $x$ is invertible, so $F$ has height exactly $m$. Hence the formal group of connected component of $F[p^\infty]$, $F[p^\infty]^0$ has height exactly $m$.
Theorem 3.2
[Lur10], Theorem 1, Lecture 14 For $F$ and $F’$ two formal group law over a ring $R$ of height exactly $h>0$, let $Iso(F,F’)$ denote the ring classifying isomorphisms from $F$ to $F’$, there is then a directed system of injective finite étale map \[ R=R(1)\to R(2)\to … \]such that $Iso(F,F’)=\colim _n R(n)$.
Definition 3.3
(I am not sure if this definition is found elsewhere, but the preprint did not give a definition, forcing me to guess what it means)
Given two formal groups of height $h$ over $R$, we know that $Iso(F,F’)$ is pro-finite-étale, a marking is then a $T$-point of $Iso(F,F’)$ for some cover $T$ over $R$, that is an isomorphism $f: F_T\to F_T’$.
There is an action of $P_n$ on $Q:=Iso(F[p^\infty]^0,\Gamma_m \otimes_k B_m)$ determined by precomposition. There is also an action of $P_m$ on $Q$ determined by postcomposition, which is transitive, determining a pro-finite-étale $P_m$-torsor over $B_m$, as $Q=\lim_h Q_h$ is pro-finite-étale, for each $Q_h:=Iso(F[p^\infty]^0,\Gamma_m \otimes_k B_m)/ U_h$ a $P_m/ U_h$-finite-étale torsor (so $Q_h$ is represented by some finite étale $B_m$-algebra. Recall that $U_h$ is the normal subgroup of $P_m$ such that $U_h$ are equivalent to the identity modulo $T^{p^{h+1}}$.
Proposition 3.3
For every $h\geq 0$, the $P_m/U_h$-torsor $Q_h$ is represented by a finite free algebra $H_{m,h}$ over $A_m$ spanned by variables $a_0,…,a_h$ such that \[ a_0^{p^m-1}=x, a_i^{p^m}-x^{p^i}a_i=P_i(a_0,…,a_{i−1}) \]for some polynomials $P_i$. That is over $B_m$, $Q_h=\spec(H_{m,h}[1/x])$. $H_{m,h}$ are freely generated by $a_0^{e_0}a_1^{e_1}…a_{h}^{e_h}$ for $0\leq e_0 <p^m-1$ and $0\leq e_i<p^m$
Moreover, the $P_n$ action on $H_{m,h}$ induced by the identification above preserves $H_{m.h}[1/x]$, and there exists a finite filtration on $F^t\subseteq F^{t-1}\subseteq …\subseteq F^0=H_{m,h}$ such that $F^{i}/F^{i+1}\cong \mathcal{L}_m^{\otimes d_i}$ for some $d_i \in \Z$. The identification and the filtration is preserved under restriction to higher height.
proof:
Let $f:F[p^\infty]^0\to \Gamma_m$ be an isomorphism over some $B_m$-algebra, which is an ordinary power series $f(T)=\sum_{j\geq 1}c_j T^j, and denote $a_i=c_{p^i}$.
We have the equality, \[ f(F[p^\infty]^0(X,Y))=\Gamma_m(f(X),f(Y)) \] Comparing coefficients and using the fact $p=0$ over $B_m$, we have that $c_i$ are determined by $a_i$, where $a_i$ are freely determined. Comparing $p$-series \[ f([p]_F T)=f(T)^{p^m} \]gives us the identity in the statement. Then we see that the generation is trivial.
Therefore, there exists a morphism $Q_h\to \spec(H_{m,h}[1/x])$ sending an isomorphism modulo $U_h$ over a $B_m$ algebra $R$ to the $R$ point corresponding to the formal power series modulo $T^{p^{h+1}}$ determined by the corresponding $R$-point of $ \spec(H_{m,h}[1/x])$.
Over separable closed fields, every height $m$ formal group law $F$ is isomorphic to $\Gamma_m$, and we have $P_m/U_h$ has rank $(p^m-1)p^{mh}$. The map $Q_h\to \spec(H_{m,h}[1/x])$ is surjective on the separable closed field(?). Therefore, surjectivity and same rank over separable closed field implies isomorphism between finite étale algebra.
The action of $G$ on $Q_h$ is determined by precomposition, that is for $g\in G$ and $f\in Q_h$, $g\cdot f\equiv f\circ \alpha_g \mod T^{p^{h+1}}$. We have that $\alpha_g(T)=\chi_g T+\sum_{l\geq 2}\beta_l T^l$, and \[ f(\alpha_g(T))=a_0\alpha_g(T)+a_1\alpha_g(T)^p+a_2\alpha_g(T)^{p^2}+…+ \text{other terms} \= g\cdot a_0 T+g\cdot a_1 T^{p}+…+ g\cdot \text{other terms} \] The only degree $1$ term in this equation is $\chi_g a_0$ hence $g(a_0)=\chi_g a_0$. Similarly, for the degree $p^i$ term, there is a coefficient of $\chi_g^{p^i} a_i$ with addition of some polynomial of coefficient $Q$, which depends on the term $a_0,…,a_{i-1}$ because all the term containing $a_{j> i}$ contribute to the term of higher degree. The coefficient of $Q$ does not contain any inverses of $x$, and $H_{m,h}$ is $x$-torsion-free, so the action of $G$ on $Q_h=\spec H_{m,h}[1/x]$ extend to an action on $H_{m,h}$.
Lastly, we impose the inverse lex ordering on $a_0^{e_0}…a_h^{e_h}$, that is comparing power $e_h$ first then break tie with $e_{h-1}$ and so on. We see the action of $g$ on $a_0^{e_0}…a_h^{e_h}$ is the following \[ g\cdot a_0^{e_0}…a_h^{e_h}=\prod_{j=0}^h\big(\chi_g^{p^j}a_j+Q_j(a_0,…,a_{j-1})\big)^{e_j} \] We define the filtration $F^e$ on $H_{m,h}$ to be generated by monomials $a^e=a_0^{e_0}…a_h^{e_h}$ by adding $1$ monomial at a time in increasing order.
The quotient of filtration $F^e/F^{e-1}$ is generated by $a_0^{e_0}…a_h^{e_h}$. Let $j$ be the greatest index such that $e_j$ is not zero. Since $a_j>Q_j(a_0,…,a_{j-1})>Q_{j-1}…$ in reverse lex order, and the only term in $g(a_0^{e_0}…a_h^{e_h})$ containing a $a_j^{e_j}$ term is $a_0^{e_0}…a_h^{e_h}$. Therefore, there is an action of $G$ on the quotient.
In $F^e/F^{e-1}$, $g(a_0^{e_0}…a_h^{e_h})=\chi_g^{e_0+pe_1+…+p^he_h}a_0^{e_0}…a_h^{e_h}$, hence \[ F^e/F^{e-1}\cong \mathcal{L}_m^{\otimes e_0+pe_1+…+p^he_h} \]
Q.E.D