Change of coordinates and derivatives in $\mathbb{R}^n$ on a boundary integral I'm am slightly confused while trying to keep everything straight between looking at integration as on a manifold vs. the diffeomorphism change of variables.
Consider a smooth domain $B\subset \mathbb{R}^2$ and the domain $B_s\subset\mathbb{R}^3$ such that $B_s=B\times[0,s]$ 
Question 1: Consider the integral
$$I=\int_{\partial B_s} f(x)\, d\sigma_{B_s}$$
where $\sigma_{B_s}$ indicates the integral over the boundary (surface measure). If we change variables, $(x_1,x_2,x_3)\to (x_1,x_2,s\xi)$, then I think we have
$$I=s\int_{\partial B\times (0,1)} f(x_1,x_2,s\xi) d\sigma_B d\xi+\int_{B\times\{0,1\}} f(x_1,x_2,s \xi)dx_1 dx_2$$
(where $d\sigma_B$ is the surface measure on $B\subset\mathbb{R}^2$)
is this correct? It bothers me that a $s$ does not comes out of the second integral although I know intuitively why (the top part has no "volume" in the third direction). I can't seem to reconcile this with the diffeomorphism version of the change of variables formula, and I would like to understand it better.
Question 2: If we considered
$$A=\int_{\partial B_s} \frac{\partial}{\partial_{x_3}} f(x) d \sigma_x,$$
we know from the chain rule that $\displaystyle\frac{\partial}{s\partial \xi}=\frac{\partial}{\partial_{x_3}}$. So when we change variables, does the problem become:
$$A=\int_{\partial B \times (0,1)}\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) s\, d\xi d\sigma_B+\int_{B \times \{0,1\}}\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,h\xi) dx_1 dx_2?$$
I am not positive how derivatives inside of the integral are affected by change of variables
It seems both of my confusions would be solved if I understood integration on manifolds better... Thank you.
 A: The formula you have in Question 1 is correct. As a sanity check, put $f\equiv 1$ in it: the result should describe the surface area.
$$A=s\int_{\partial B\times (0,1)} d\sigma_B\, d\xi+\int_{B\times\{0,1\}} dx_1\, dx_2 = 
s\int_{\partial B} d\sigma_B+\int_{B\times\{0,1\}} dx_1\, dx_2 $$
Now the fact that $s$ is present only in front of the first integral makes  perfect geometric sense: stretching in vertical direction leaves the areas of top and bottom surfaces unchanged, while the side area is multiplied by the stretch factor. 
Concerning question 2: to minimize confusion, it helps to use the notation $f_3(x_1,x_2,x_3)$, where subscript indicates differentiation in the third variable (whatever it is called). Since $f_3$ is just an ordinary function, the formula from Question 1 applies to it:
$$\int_{\partial B_s} f_3(x)\, d\sigma_{B_s} = s\int_{\partial B\times (0,1)} f_3(x_1,x_2,s\xi) d\sigma_B d\xi+\int_{B\times\{0,1\}} f_3(x_1,x_2,s \xi)dx_1 dx_2 \tag1$$
The next step is to relate $f_3(x_1,x_2,s\xi)$ and $\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi)$. This is what the chain rule does: 
$$\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) = f_3(x_1,x_2,s\xi)\, \frac{\partial(s\xi)}{\partial \xi} =f_3(x_1,x_2,s\xi)\,s $$
Therefore, $f_3(x_1,x_2,s\xi)$ can be replaced by $\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi)$ everywhere in (1): 
$$\int_{\partial B_s} \frac{\partial }{\partial x_3}f(x)\, d\sigma_{B_s} = \int_{\partial B\times (0,1)} \frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) d\sigma_B d\xi + \frac{1}{s}\int_{B\times\{0,1\}} \frac{\partial}{\partial \xi}f(x_1,x_2,s \xi)dx_1 dx_2 $$

By the way, it's a good idea to write $\frac{\partial }{\partial \xi} f(x_1,x_2,s\xi)$ (as you have done) instead of $\frac{\partial f}{\partial \xi} (x_1,x_2,s\xi)$. I find the latter notation a frequent source of confusion in calculations like these - do we take derivative first and rescale later, or the other way around? 
