How does the problem of having a coffee-machine close relate to vertex cover?
As I already pointed out in my comment, there is in fact a need for a transformation of the problem. The problem you ask for is not known as vertex cover, but as dominating set.
The proof then goes as follows:

We transform an instance of vertex cover $(V, E, k)$ as follows (we assume that the graph $(V,E)$ is connected). For every edge - just as the hint says - add two edges and one vertex. More precisely, for an edge $uv \in E$, add a new node $w \not\in V$ and edges $vw, wu$ to the graph. The resulting graph constitutes an instance of Coffee in the obvious way, that is, the "close-enough"-relation is represented by edges and the meeting rooms are represented by vertices. Let $(V', E', k)$ denote the instance of coffee produced by the transformation.

We now need to show now that $(V, E)$ has a vertex cover of size $\leq k$ if and only if for $(V', E')$ it suffices to use $\leq k$ coffee machines.

"$\Rightarrow$": Let $v_1, \ldots, v_l, l \leq k$ be a vertex cover of $(V,E)$. It is clear that the same choice of nodes produces a valid choice for coffee machine placement on $(V', E')$, as we assumed $(V, E)$ to be connected.

"$\Leftarrow$": Let $v_1, \ldots, v_l, l \leq k$ be a valid choice for coffee machine placement. WLOG, assume that $\forall i \in \{1, \ldots, l\}: v_i \in V$, for if $v_s \in V' - V$ with $1 \leq s \leq l$, we can obtain another valid choice for coffee machine placement by replacing $v_s$ with $v \in V$ such that $vv_s \in E' - E$, that is, with one of the two nodes "between" which $v_s$ was inserted by the transformation. Note that these two nodes were also precisely the ones that were covered by choosing $v_s$ to have a coffee machine. By construction, there must have been an edge between the two nodes and an edge between $v_s$ and each of the two nodes. So by replacing $v_s$ with one of them, we don't shrink the set of covered nodes. 
Then, $v_1, \ldots, v_l$ constitutes a vertex cover of $(V, E)$. To show this, let $uv \in E$. By construction, there is $w \in V' - V$ and $uw, wv \in E' - E$, and no other edges are incident to $w$. As $(V', E')$ covers $w$, one of $u$ or $v$ need must have been chosen as places for a coffee machine, and thus, $uv$ is incident to at least one node in $\{v_1, \ldots, v_l\}$. Thus, $v_1, \ldots, v_l$ is a vertex cover of size $l \leq k$.