The mean value theorem of harmonic function tells us: If $U \subset \Bbb R^n$ open, $f: U \to \Bbb R$ harmonic then for all $y \in U$ and $r>0$ s.t. $\overline{B}(y,r) \subset U$ the following is true: $$ f(y) = \frac{1}{vol_{n-1}( \partial B(yr))}\int_{\partial B(y,r)}f d vol_{n-1}$$
Now the proof starts with the following observation: $vol_{n-1}(\partial B(y,r))= n \cdot w_n \cdot r^{n-1}$, where $w_n:=\mu(B^n(0,1))$ (the jordan measure of the unit disc). First how we can see this?
The next point is a strange integral transformation:
$$\frac{1}{S^{n-1}}\int_{\partial B(y,r)} f d vol_{n-1}=\int_{S^{n-1}}f(y+rw)d vol_{n-1}(w)$$ How can we conclude this?
Many thanks for some help!
Hint
For your first question, use polar coordinates.
For your second question, first of all, I guess that $\frac{1}{S^{n-1}}$ is wrong. It should be $\frac{1}{|\partial B(y,r)|}$, and on the RHS, a $\frac{1}{|S^{n-1}|}$ is missing (or something like that). Then, $$\partial B(y,r)=\{y+r\sigma \mid \sigma \in S^{n-1}\},$$ and $$\mathrm d z=d(y+r\sigma )=r^{n-1}\,\mathrm d \sigma,$$ where $\mathrm d z$ is an area element of $\partial B(y,r)$ and $\mathrm d \sigma $ an area element of $S^{n-1}$.