By definition the open set $D\subset\mathbb R^n$ is said to have differentiable boundary $\partial D$ of class $C^k$ ath the point $p\in \partial D$ if there is a neighborhood $U$ of $p$ and a real valued function $r\in C^k(U)$ with the following properties: $$U\cap D=\{x\in U:r(x)<0\};\\dr(x)\not=0 \ \ \ \text{ for } x\in U.$$
Now how does this imply that $U\cap\partial D=\{x\in U:r(x)=0\}$ and $U-\bar D=\{x\in U:r(x)>0\}$?
First, suppose that $r(x)<0$ for some $x \in U\cap\partial D$. Since $r\in C^k(U)$, we can then find a neighborhood $V$ of $x$ so that $|r(x)-r(y)|<\frac{1}{2}|r(x)|$ for all $y\in V$. In particular, $r(y)<0$, so $y\in D$. This means that $V\cap D$ is an open subset of $D$ (in the subset topology) containing $x$, meaning that $x$ is an interior point of $D$. This contradicts our assumption that $x\in\partial D$, so our assumption that $r(x)<0$ must be false. Similarly, show that if $r(x)>0$, then $x$ is an interior point of the complement $D^c$. This means that we must have $r(x)=0$.
Edit: I realize now that since we assumed $D$ to be open, we can rule out $r(x)<0$ even faster. If $r(x)<0$, then $x \in U\cap D$ and is thus an interior point of $D$. So we must have $r(x)\geq 0$ for all $x \in U\cap \partial D$. Now rule out $r(x)>0$ using the argument above.
For the inclusion $U\cap\partial D\supset\{x\in U:r(x)=0\}$, use the fact that $dr(x)\neq 0$ for $x\in U$ to show that if $r(x)=0$, then every neighborhood of $x$ has points at which $r$ is positive and points at which $r$ is negative. This tells you that every neighborhood of $x$ intersects both $D$ and $D^c$, and thus that $x$ is a boundary point of $D$.
Once you've established that $U\cap\partial D=\{x\in U:r(x)=0\}$, the fact that $U-\overline{D}=\{x\in U:r(x)>0\}$ will follow from the fact that $\overline{D}=D\cup\partial D$.
Follow up to comment: Let $x\in U$ be such that $r(x)=0$, and let $V\subset U$ be an arbitrary neighborhood of $x$. Since $dr(x)$ is a nonzero linear map, choose $y$ so that $dr(x)(y)<0$. We know that \begin{equation} dr(x)(y) = \lim_{h\to 0} \frac{r(x+hy)-r(x)}{h}, \end{equation} so let $\epsilon=\frac{1}{2}|dr(x)(y)|$ and choose $\delta>0$ so that if $h<\delta$ then \begin{equation} \left\vert dr(x)(y) - \frac{r(x+hy)-r(x)}{h}\right\vert < \epsilon. \end{equation} Using the fact that $r(x)=0$ and our choice of $\epsilon$, you can rearrange this to conclude that \begin{equation} r(x+hy) < \frac{1}{2}h\cdot dr(x)(y) < 0. \end{equation} Now choose $h>0$ small enough that $h<\delta$ and $x+hy\in V$. From this we see that every neighborhood of $x$ intersects the set $\{y\in U:r(y)<0\}=U\cap D$, and thus that $x$ is in the boundary of this set.