By Taylor's theorem with the integral form of the remainder, \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|x|^3/6 \tag{1} \end{equation*} and \begin{equation*} |e^{ix}-1-ix|\le x^2/2 \end{equation*} for real $x$. The latter inequality also implies \begin{equation*} |e^{ix}-1-ix-(ix)^2/2|\le|e^{ix}-1-ix|+|(ix)^2/2|\le x^2/2+x^2/2=x^2. \tag{2} \end{equation*}
Let now $X:=X_{n,k}$, $F:=F_{n,k-1}$, $E_F Z:=E(Z|F)$, and $c:=\epsilon$. We need to show that $$|E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2|\le \tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c} $$ given that $E_F X=0$.
Let $r(x):=e^{ix}-1-ix-(ix)^2/2$ and write, in view of (1) and (2): \begin{multline*} |E_F\exp(itX)-1-\tfrac12\,t^2E_F X^2| =|E_F r(itX)|\le|E_F r(itX)\,1_{|X|\le c}|+|E_F r(itX)\,1_{|X|<c}| \\ \le\tfrac16\,|t|^3 E_F|X|^3\,1_{|X|\le c}+t^2E_FX^2 1_{|X|>c}, \end{multline*} as desired.