$\newcommand{\Ex}{\mathbb E}$ Let $X^i, X^j, X^k$ be independent copies of a stochastic process $X$. Let $\beta:\mathbb R \to \mathbb R$ be bounded continuous. Let $b(s, x) := \Ex[\beta(x- X_s)]$ for all $s \in [0, \infty)$ and $x \in \mathbb R$. Let $$ \varphi (s) :=\Ex [(\beta(X^i_s - X^j_s) - b(s, X^i_s)) (\beta(X^i_s - X^k_s) - b(s, X^i_s))] \quad \forall s \in [0, \infty). $$
Then it's mentioned in the proof of Lemma 5.4. of this paper that
$\varphi (s)=0$ for all $s \in [0, \infty)$.
I decompose $$ \begin{align} \varphi (s) &= \Ex [\beta(X^i_s - X^j_s) \beta(X^i_s - X^k_s)] - \Ex [\beta(X^i_s - X^j_s) b(s, X^i_s)] \\ & \qquad - \Ex [\beta(X^i_s - X^k_s) b(s, X^i_s)] + \Ex [b^2(s, X^i_s)], \end{align} $$ but I could not see how $\varphi (s)=0$. Could you elaborate on this point?
I have not read the paper, but the following may be useful to you. All equalities with conditional expectations are in the a.e. sense.
Let $X,Y,Z$ be real-valued random variables. Suppose $P(X \in A,Y \in B|Z)=P(X \in A|Z)P(Y \in B|Z)$ for all $A,B \in \mathscr{B}(\mathbb{R})$ (i.e. we have conditional independence of $X$ and $Y$ wrt $Z$). Let $A,B,C,D \in \mathscr{B}(\mathbb{R})$. Then $$\begin{aligned}E[\mathbf{1}_{A \times B}(X,Z)\mathbf{1}_{C \times D}(Y,Z)|Z]&=E[\mathbf{1}_A(X)\mathbf{1}_{C}(Y)|Z]\mathbf{1}_{B}(Z)\mathbf{1}_{D}(Z)=\\ &=E[\mathbf{1}_A(X)|Z]E[\mathbf{1}_{C}(Y)|Z]\mathbf{1}_{B}(Z)\mathbf{1}_{D}(Z)=\\ &=E[\mathbf{1}_{A \times B}(X,Z)|Z]E[\mathbf{1}_{C \times D}(Y,Z)|Z]\end{aligned}$$ So for $C,D$ fixed we have $$\mathscr{B}(\mathbb{R})\times \mathscr{B}(\mathbb{R})\subseteq \{F \in \mathscr{B}(\mathbb{R}^2):E[\mathbf{1}_F(X,Z)\mathbf{1}_{C\times D}(Y,Z)|Z]=E[\mathbf{1}_{F}(X,Z)|Z]E[\mathbf{1}_{C \times D}(Y,Z)|Z]\}$$ The lhs is a $\pi$-system generating $\mathscr{B}(\mathbb{R}^2)$. The rhs is a $\lambda$-system. So $$E[\mathbf{1}_F(X,Z)\mathbf{1}_{C\times D}(Y,Z)|Z]=E[\mathbf{1}_{F}(X,Z)|Z]E[\mathbf{1}_{C \times D}(Y,Z)|Z],\,\forall F \in \mathscr{B}(\mathbb{R}^2)$$ Since $C,D$ were arbitrary, an equivalent argument shows that $$E[\mathbf{1}_F(X,Z)\mathbf{1}_{G}(Y,Z)|Z]=E[\mathbf{1}_{F}(X,Z)|Z]E[\mathbf{1}_{G}(Y,Z)|Z],\,\forall F,G \in \mathscr{B}(\mathbb{R}^2)$$ Now let $f,g$ be bounded measurable functions. A standard argument with simple functions and convergence theorems shows that $$E[f(X,Z)g(Y,Z)|Z]=E[f(X,Z)|Z]E[g(Y,Z)|Z]$$
Now suppose $(X,Y)$ is independent from $Z$ and $X$ is independent from $Y$. Then for $A,B \in \mathscr{B}(\mathbb{R})$ $$\begin{aligned}P(X \in A,Y \in B|Z)&=P(X \in A,Y \in B)=\\ &=P(X \in A)P(Y \in B)=\\ &=P(X \in A|Z)P(Y \in B|Z) \end{aligned}$$ so the above holds.
Recall that if $X$ is independent from $Z$, $Y$ is independent from $Z$ and $X \sim Y$ (i.e. they have identical distribution) then for bounded measurable $f$ we have $$E[f(X,Z)|Z]=E[f(X,z)]|_{z=Z}=E[f(Y,z)]|_{z=Z}=E[f(Y,Z)|Z]$$ We may define $b(z):=E[f(X,z)]=E[f(Y,z)]$; note: $b(Z)=E[f(X,Z)|Z]=E[f(Y,Z)|Z]$.
Finally, suppose $(X,Y)$ is independent from $Z$ and $X$ is independent from $Y$ and $X \sim Y$ and $b(z):=E[f(X,z)]=E[f(Y,z)]$ for a bounded measurable $f$. Then $$\begin{aligned}E[(f(X,Z)-b(Z))(f(Y,Z)-b(Z))]&=E[f(X,Z)f(Y,Z)]+\\ &-E[f(X,Z)b(Z)]+\\ &-E[f(Y,Z)b(Z)]+\\ &+E[b(Z)^2]=\\ &=E[E[f(X,Z)f(Y,Z)|Z]]+\\ &-E[E[f(X,Z)|Z]b(Z)]+\\ &-E[E[f(Y,Z)|Z]b(Z)]+\\ &+E[b(Z)^2]=\\ &=E[E[f(X,Z)|Z]E[f(Y,Z)|Z]]+\\ &-E[b(Z)^2]+\\ &-E[b(Z)^2]+\\ &+E[b(Z)^2]=0 \end{aligned}$$