Denote $X=(X_1,\dots,X_n)\sim \mu$ and $Y=X_{n+1}\sim \nu$, then $X,Y$ are independent and $(X,Y)\sim \mu \times \nu$. Also,
$$ g(x)=\int_{\mathbb{R}}h(x,y)\nu(dy) \quad x\in \mathbb{R}^n.$$ Thus $$g(X)=\mathbb{E}[h(X,Y)|X].$$
In fact, $g(X)$ is $\sigma(X)$-measurable and
$$ \mathbb{E}[g(X)1_{\{X\in A\}}]=
\mathbb{E}[g(X)1_A(X)]= \int_{\mathbb{R}^n} g(x)1_A(x) \mu(dx)=\int_{\mathbb{R}^n} \left( \int_{\mathbb{R}} h(x,y)\nu(dy)\right)1_A(x)\mu(dx) = \int_{\mathbb{R}^n\times \mathbb{R}} h(x,y)1_A(x)\mu\times \nu(dx\times dy) = \mathbb{E}[h(X,Y)1_A(X)]=
\mathbb{E}[h(X,Y)1_{\{X\in A\}}]$$ for all $A\in \mathcal{B}_{\mathbb{R}^n}$.
What Klenke says is true and can be verified by computing conditional densities. But I will suggest general direct approaches to compute both the distribution of $(X_{(n)}, \dots, X_{(1)})$ and of $Z = (Y_1, Y_1 + Y_2, \dots, Y_1 + \dots + Y_n)$ when $X_1, \dots, X_n$ are arbitrary i.i.d. rvs with densities, and $Y_1, \dots, Y_n$ are arbitrary indepndent rvs with densities, and then you can specialize to this specific problem.
It is a consequence of the change of variables theorem that whenever $X_1, \dots, X_n$ are i.i.d. and have density $f_{X_1}$ (wrt Lebesgue measure), the density $f$ of $(X_{(n)}, \dots, X_{(1)})$ is
$$f(x_n, \dots, x_1) = n!f_{X_1}(x_n) \dots f_{X_1}(x_1), \hspace{20pt} x_n < x_{n - 1} < \dots < x_1.$$
Suppose $Y_1, \dots, Y_n$ are independent and that each $Y_j$ has density $f_{Y_j}$ wrt Lebesgue measure. We have $Z = AY$, where $A \in GL(n, \mathbb{R})$ and $\det(A) = 1$. So by the change of variables theorem of integration, $Z$ has density
$$f_Z(z) = f_Y(y)|\det(A)|^{-1} = f_Y(y), \hspace{20pt} y = A^{-1}z.$$
Edit:
To verify Klenke's claim about $\mathcal{L}(X_{(1)} - X_{(n)}, \dots, X_{(n-1)} - X_{(n)} \mid X_{(n)})$, fix $x_n \in \mathbb{R}$. We need to compute the measure
$$\mathcal{L}(X_{(1)} - X_{(n)}, \dots, X_{(n-1)} - X_{(n)} \mid X_{(n)} = x_n) = \mathcal{L}(X_{(1)} - x_n, \dots, X_{(n-1)} - x_n \mid X_{(n)} = x_n).$$
Assuming the joint density $f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}$ exists (which will be proven below), we get that $\mathcal{L}(X_{(1)} - x_n, \dots, X_{(n-1)} - x_n \mid X_{(n)} = x_n)$ has density
$$f(x_1, \dots, x_{n - 1}) = f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}(x_1, \dots, x_n)/f_{X_{(n)}}(x_n).$$
The existence and formula for the joint density $f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}$ follows from the change of variables theorem as above. The rest is plugging in the formulas for when $X_i$ are i.i.d. exponential(1).
Best Answer
$$ E[X_1\wedge X_2|X_1=s]=\int_0^\infty (t\wedge s)\hspace{-.5cm}\underbrace{\theta e^{-\theta t}}_{\text{conditional pdf}\\\text{of $X_2$ given $X_1=s$}}\hspace{-.5cm}\,dt=\int_0^st\theta e^{-t\theta}\,dt+\int_s^\infty s\theta e^{-t\theta}\,dt $$