There is a general concept which incorporates both polyhedra and cones, namely the concept of a convex body. And the concept of a vertex can be defined in that general setting.
By definition, a subset $C \subset \mathbb R^3$ is a convex body if there exists a set of half-spaces $\{H_i\}_{i \in I}$ such that
$$C = \bigcap_{i \in I} H_i
$$
And, to be complete, a (closed) half space $H$ is the solution set of an equation of the form $ax + by + cz + d \ge 0$, for some constants $a,b,c,d \in \mathbb R$ such that at least one of $a,b,c$ is nonzero:
$$H = \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d \ge 0\}
$$
Some extra terminology: the boundary plane of $H$ is
$$\partial H = \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d =0\}
$$
If $H$ is a half-space, if $C \subset H$, and if $C \cap \partial H \ne \emptyset$, then we say that $H$ is a supporting half-space of $C$. In this terminology, $C$ is equal to the intersection of the set of all supporting half-spaces of $C$.
So, for example, a finite sided (convex) 3-D polyhedron can be defined as a convex body $C \subset \mathbb R^3$ which is the intersection of a finite set of closed half spaces (this allows for finite sided polyhedra to be unbounded; one could also restrict ones attention just to finite sided polyhedra that are bounded).
Let me also define a cone in this setting, more specifically a right circular cone (this is just to keep things simple; one could study more general cones). First, one is given a 2-dimensional plane $P \subset \mathbb R^3$, a line $L \subset \mathbb R^3$ perpendicular to $P$, a point $Q = P \cap L$, another point $R \in L - \{Q\}$, and a radius $s > 0$. One takes $C$ to be the circle in $P$ of radius $s$ centered on $Q$.
For each $x \in C$ one takes $H_x$ to be the half space such that its boundary plane $\partial H_x$ passes through $R$ and intersects the plane $P$ in a line that is tangent to $C$ at the point $x$, and such that the half space $H_x$ entirely contains the circle $C$.
With that general definition and these two special examples, here's the definition of a vertex. First, given a cone $C \subset \mathbb R^3$ and a point $p \in c$, to say that $p$ is an extreme point of $C$ means that there exists a closed half-space $H \subset \mathbb R^3$ such that $H \cap C = \{p\}$ (this implies that $p$ is contained in the boundary plane $\partial H$; this also implies that the "opposing" half space of $H$ is a supporting half-space of $C$). Let's also say that any half-space $H$ satisfying the above property is a witness to the extremity of $p$.
A half space that witnesses the extremity of $p \in C$ might not be unique, and we need to quantify the non-uniqueness. For each half-space $H$ that witnesses the extremity of $p$, let $v^\perp_H$ be the unique unit vector based at $p$ which is orthogonal to $\partial H$ and points into $H$. The half-space $H$ determines and is determined by the vector $v^\perp_H$, so I'll extend the terminology to say that the vector $v^\perp_H$ witnesses the extremity of $p$.
To say that an extreme point $p$ is a vertex means that there exist three non-coplanar vectors $v^\perp_1,v^\perp_2,v^\perp_3$ each based at $p$, and each witnessing the extremity of $p$.
The unique vertex of the right circular cone defined above is the point $R$. Also, in a finite sided polyhedron, a point is a vertex if and only if it is a common endpoint of a set of three edges. (This begs the definition of edges. First, let's say that an extreme point $x \in C$ is an edge point if the set of vectors $v_H$ which witness the extremity of $x$ traces out a circular arc in the unit sphere centered on $x$; and then let's say that an edge is a set of edge points all of which share the same set of witnessing vectors.)
One other interesting example is the 3-dimensional ball $B^3 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^3 \le 1\}$, which is a convex body. It's set of extreme points is the unit sphere $S^2 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 = 1\}$, but $B^3$ has no vertices and no edges, because the witnessing half-space of each extreme point is unique.
I think the thing that you're interested in here is the covering density of the sphere, rather than its packing density - this is the lowest-density arrangement of overlapping unit spheres needed to cover every point in $\mathbb{R}^3$.
In the plane, the hexagonal lattice is optimal for both packing and covering. In three dimensions, though, they differ - the FCC lattice is best for packing, but the BCC lattice (a cubic lattice along with the centers of each cube) is the lattice offering the thinnest covering density of $\frac{5\sqrt{5}\pi}{24}\approx1.4635$. (Whether a non-lattice packing could do better is an open problem, I believe - see here.)
If you take the Voronoi cells of the BCC lattice with side length $1$, you get truncated octahedra with edge length $\sqrt{2}/4$, volume $1/2$, and circumradius $\sqrt{5}/4$, thus taking up $\frac{24}{5\sqrt5\pi} \approx 0.6833$ of the sphere, which of course ends up being the reciprocal of the covering density.
This correspondence between covering density of a sphere arrangement and relative volume of the voronoi cell means that the truncated octahedron is optimal, at least among space-filling bodies which tile by translation, and that any improvement would lead to a previously-unknown thinnest sphere covering. (The converse is not necessarily true, since the Voronoi cells of a non-lattice packing don't have to be congruent.)
For reference, though, a rhombic dodecahedron with coordinates $(\pm1,\pm1,\pm1)$ and all permutations of $(\pm2,0,0)$ has volume $16$ within a sphere of radius $2$, so occupies only $\frac{3}{2\pi}\approx0.4775$ of the circumsphere.
Best Answer
This is supposed to be a comment but I would like to post a picture.
For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere
$$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$
Their convex hull will be a $m$-gonal bipyramid which appear below.
Up to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.
Let $\phi = \cos^{-1}\sqrt{\frac{15+\sqrt{145}}{40}}$, one possible set of vertices are given below: $$ ( \pm \sin3\phi, 0, +\cos3\phi ),\;\; ( \pm\sin\phi, 0,+\cos\phi ),\\ (0, \pm\sin3\phi, -\cos3\phi),\;\; ( 0, \pm\sin\phi, -\cos\phi). $$ For this set of vertices, the polyhedron is the convex hull of two polylines. One in $xz$-plane and the other in $yz$-plane. Following is a figure of this polyhedron, the red/green/blue arrows are the $x/y/z$-axes respectively.
$\hspace0.75in$
For $n \le 8$, above configurations are known to be optimal. A proof can be found in the paper
An online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).
For $n \le 130$, a good source of close to optimal configurations can be found under N.J.A. Sloane's web page on Maximal Volume Spherical Codes. It contains the best known configuration at least up to year 1994. For example, you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8 files under the link to library of 3-d arrangements there.