To parametrize a $(R,r)-\textbf{Torus}$ where $r<R$, we revolve a circle of radius $R$ in the $xz$-plane, about the $z$-axis. One can come up with a parametrization by taking:
$$\Phi(t,s) = \gamma(s)+ r\bigg(\textbf{n}(s) \cos (t) + \textbf{b}(s) \sin(t)\bigg)$$
where $\gamma(s) = \left(0,R \cos \left(\frac{s}{R}\right), R \sin \left(\frac{s}{R}\right)\right)$. The above map simply wraps a tube of radius $r$ about the circle of radius $R$ in the $z = 0$ plane. Simplifying the above we have that:
$$\Phi(t,s) = ((R+r \cos(t)) \cos (s), (R+r \cos(t)) \sin(s), R \sin(t)) = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$
Here you can take $(t,s) \in U=(0,2\pi) \times (0,2\pi)$ so that $\Phi$ is injective. Observe that we've missed various points, namely those in the image of $\Phi(t,\pi/2)$ and $\Phi(0,s)$.
Now just take another restriction of the plane, i.e $(t,s) \in V=(\epsilon,\epsilon+2\pi) \times (\epsilon,\epsilon+2\pi)$ where $\epsilon$ is small. Hence we have:
$$\{(\Phi|_U,U),(\Phi|_V,V)\}$$
is an atlas on $T^2$ (where you are left to check the other conditions). Use this same method to show that the cylinder is a surface.
If $S$ was orientable, $S$ would have an atlas $\mathcal T = (T_j, \varphi_j)$ where the Jacobians of all the transition functions are positive.
Denote $C_1,C_2$ the two connected components of $U \cap V$ and take $u_i \in T_i \cap C_i$ for $i \in \{1,2\}$. As $S$ is path connected, consider a path $f$ joining $u_1$ to $u_2$. A chain of charts of $\mathcal T$ covering $f$ would have all positive transition functions. A contradiction with the hypothesis regarding the $U,V$ atlas.
Best Answer
Let $\varphi :D\to R\subseteq S$ be the given homeomorphism and suppose by contradiction that $x_0$ is a point in the boundary of $D$ such that $\varphi (x_0)$ lies in the interior of $R$.
Pick a neighborhood $U$ of $\varphi (x_0)$ homeomorphic to an open disc (e.g. a coordinate neighborhood), and contained in $R$. Letting $V=\varphi ^{-1}(U)$ we have that $V$ is also homeomorphic to an open disc.
Consequently $V\setminus \{x_0\}$ is homeomorphic to a punctured disk and hence not simply conected. This is a contradiction because every open simply connected subset of $D$ remains simply connected once a point of the boundary of $D$ is removed from it.