If you take expectation first with respect to $S_{n-1}$, then by Fubini's theorem the last term gives $$E \left[\frac{|X_n|}{\sqrt{n}}\frac{1}{\lambda} \int_0^1 P\left(z-t\frac{X_n}{\sqrt{n}} \le S_{n-1} \le z-t\frac{X_n}{\sqrt{n}} + \lambda\right) dt\right] dt\right].$$ Now if $Y$ is a standard Gaussian random variable and $a\in \mathbb{R}$, then $$P(a\le S_{n-1} \le a+\lambda) \le P(a\le Y \left(E\frac{|X_n|}{\sqrt{n}} le a+\lambda) + 2\delta(\gamma,n-1) \right)\frac{2}{\lambda} le \delta(\gamma,n-1). frac{\lambda}{\sqrt{2\pi}} + 2\delta(\gamma,n-1),$$ so the expectation above is bounded by $\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{2\pi}}+2\frac{\delta(\gamma,n-1)}{\lambda}\right)$.
If you take expectation first with respect to $S_{n-1}$, then by Fubini's theorem the last term gives $$E \left[\frac{|X_n|}{\sqrt{n}}\frac{1}{\lambda} \int_0^1 P\left(z-t\frac{X_n}{\sqrt{n}} \le S_{n-1} \le z-t\frac{X_n}{\sqrt{n}} + \lambda\right) dt\right] \le \left(E\frac{|X_n|}{\sqrt{n}} \right)\frac{2}{\lambda} \delta(\gamma,n-1).$$