First part of your question
Basically it's all to do with the inverse function theorem. But more explicitly:
Let $\mathbf{x}:U \subset \mathbb{R}^2 \rightarrow S$ be a parameterization and $q$ be a point in $U$.
Define $\mathbf{F}:U \times I \rightarrow \mathbb{R}^3$ by
$$
\mathbf{F}(u,v,t):= \mathbf{x}(u,v) + t\cdot \hat{e}_3, \hspace{1in} t \in I.
$$
Another way to think about $\mathbf{F}$ is as follows:
If we know that $\mathbf{x}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$, then
$$
\mathbf{F}(u,v,t) = \langle x(u,v), y(u,v), z(u,v) +t \rangle.
$$
So, given that $\mathbf{x}(u,v)$ is a parameterization, we know that it must be differentiable, thus $\mathbf{F}(u,v,t)$ is also differentiable.
Furthermore, since $\mathbf{x}(u,v)$ is a parameterization, we can assume without loss of generality, that
$$
\dfrac{\partial(x,y)}{\partial(u,v)}(q) \not= 0.
$$
Now take the differential of $\mathbf{F}(u,v,t)$ at $q$:
$$
d\mathbf{F}_q = \left| \begin{array}{ccc}
x_u & x_v & 0 \\
y_u & y_v & 0 \\
z_u & z_v & 1 \end{array} \right|_{q} = \dfrac{\partial(x,y)}{\partial(u,v)}(q).
$$
Thus $d\mathbf{F}_q \not=0$. This satisfies the condition for the inverse function theorem. Hence, there exists an open neighborhood, say $M$, around $\mathbf{F}(q) \in \mathbb{R}^3$ such that $\mathbf{F}^{-1}$ exists and is differentiable.
But if we set $t=0$ in the expression for $\mathbf{F}(u,v,t)$, then $\mathbf{x}(u,v) = \mathbf{F}(u,v,0)$. Hence $\mathbf{x}^{-1}$ is differentiable around $q$. Since $q$ was arbitrarily chosen, we know that this holds for all points in $U$. Hence $\mathbf{x}^{-1}:\mathbf{x}(U) \rightarrow R^2$ is differentiable.
Second part of your question
If there exists a diffeomorphism between $U$ and $\mathbf{x}(U)$, then $U$ and $\mathbf{x}(U)$ are diffeomorphic. But we already know that $\mathbf{x}(U)$ is differentiable and invertible by hypothesis; and we just showed that $\mathbf{x}^{-1}$ exists and is differentiable. Thus $\mathbf{x}(u,v)$ is a diffeomorphism, so $U$ and $\mathbf{x}(U)$ are diffeomorphic to each other.
Remarks
I like to think about $U$ as a sheet of rubbery paper, and $\mathbf{x}(U)$ as me bending the paper to look like the surface. Then $\mathbf{x}^{-1}$ is just me bending the paper back into its' original shape. Moreover, you want to make sure that your logic does not depend on this special sheet of paper; so you take another rubbery sheet of paper, $V$, and bend that into the shape of the surface as well by $\mathbf{y}(V)$. Now I like to think that if I can bend the original rubbery sheet of paper, $U$, into the other rubbery sheet of paper, $V$, then I have all my bases covered and my logic does not depend on any particular sheet. The Change of Parameters theorem basically tells us that you can bend $U$ into $V$ by $\mathbf{y}^{-1} \circ \mathbf{x}: U \rightarrow V$ or vice-versa if you want to bend $V$ into $U$ by $\mathbf{x}^{-1} \circ \mathbf{y}: V \rightarrow U$.
Best Answer
Quasi-Definition. Given a function $F:U \subset \mathbb R^n \to \mathbb R^m$ ($U$ an open set of $\mathbb R^n$), we say that $p \in U$ is a critical value of $F$ if all its partial derivates vanishes at $p$, and $F(p)$ is called a critical value of F. A point of $\mathbb R^m$ wich is not a critical value is called a regular value of $F$.
Proposition. If $f:U\subset \mathbb R^3 \to \mathbb R$ is a differentiable function and $a\in f(U)$ is a regular value of $f$, then $f^{-1}(a)$ is a regular surface in $\mathbb R^3$.
Lets take $f:M \to \mathbb R$ with $f(x,y,z)=x^2+y^2-r^2$. First of all $f$ is differentiable, so we can calculate its critical points. So $$\frac{\partial f}{\partial x}=2x$$ $$\frac{\partial f}{\partial y}=2y$$ $$\frac{\partial f}{\partial z}=0$$ now the critical points of $f$ are of the form $(0,0,z), z\in \mathbb R$.Then $f(0,0,z)=-r^2 \neq 0$. So 0 is a regular value of $f$ and the point $(0,0,z) \notin f^{-1}(0)$. But notice that $f^{-1}(0)=M$ it follows that $m$ is a regular surface. For the second part you need to prove that $x$ is differentiable and one-to-one, and $\frac{\partial f}{\partial x} \times \frac{\partial f}{\partial y}$ is nowhere zero. I'm don't know the aswer. But if we define $U=(0,2\pi)\times \mathbb R$ then $x:U \to \mathbb R^3$ covers M except the rect line $x=r$ so we need another paremetrization, let $x':U \to \mathbb R^3$ with $x'(u,v)=(r\sin u,r\cos u, v)$ this one covers M except the rect line $y=r$.So the part that $x$ do not cover, $x'$ cover it, and viceversa.