All manifolds are smooth.
Let $M = G \cup K$. The interior of $M$ is an open set in $\mathbb{R}^n$ and can be given a global coordinate by the identity map. The points in $M$ not on the interior of $M$ are necessarily elements of $G$ and already have coordinate maps to open subsets of $\mathbb{R}^{n-1}$.
How do I give a manifold structure to the points on the boundary of $M$?
I would like an answer that doesn't (explicitly) mention "tangent spaces" because I have reason to believe that a more elementary solution exists.
I assume, the $(n-1)$-manifold $G$ is smoothly embedded in $\Bbb R^n$ and is identified with its image therein (being just the top. boundary of the open set $K$).
Yes, let $x$ be on the topological boundary of $K$, then necessarily $x\in G$.
Hint: Let the local coordinate map be denoted by $\varphi:U\to W$, with $U\subseteq G\subseteq\Bbb R^n$ and $W\subseteq\Bbb R^{n-1}$ open. This is a diffeomorphism, and we can talk about the differential of $\varphi^{-1}$ as of a smooth function of type $\Bbb R^{n-1}\to\Bbb R^n$. Thus we can take the tangent vectors at $x\in G\subseteq\Bbb R^n$, which form an $n-1$ dimensional linear subspace (if transated to the origin), so we can choose any other vector $v$, and then consider $$(w,t)\ \mapsto \ \varphi^{-1}(w)+tv\,.$$