Let me say something about the case for polyhedra (it may generalize, but I am not sure about the details).
It is not too hard to imagine that there might be a "combinatorial duality" for spherical polyhedra, in the sense that the dual of a spherical polyhedron exists, but is only determined up to combinatorial equivalence (e.g. via dual planar graphs).
But I want to argue that there can not be a geometric duality, i.e. a duality that to every concrete spherical polyhedron gives you another one, and taking the dual again brings you back to the original one.
The reason is, that given the combinatorial type of a spherical polyedron, the realization space of that type (i.e. the space of all spherical polyhedra with this combinatorial type) has a local dimension of $2n$, where $n$ is the number of vertices.
What do I mean by that: you can describe your spherical polyhedron basically by drawing some points on the sphere, and stating between which points there should be a line. The line is then uniquely determined as the great circle arc between these points (yes, there is a choice which arc to take, but lets ignore this for now).
So if we placed our points carefully, then none of these arcs intersect, and what we have is a spherical polyhedron.
But note that we can move each point slighly, and the arcs move accordingly. And if we moved the points slightly enough, then the arcs stay disjoint, and the construct stays a spherical polyhedron.
Since each vertex moves on the surface of the 2-sphere, each vertex has two degrees of freedom, and the whole construct has $2n$ degrees of freedom.
Now, consider the spherical cube, whose dual (if our duality is meaningful in any way) is the spherical octahedron.
But the first one has $2\times 8=16$ degrees of freedom, and the latter one only $2\times 6=12$.
So not every unique realization of the spherical cube can be mapped into a unique realization of the spherical octahedron, and so the geometric duality fails.
Yes, in general this is how one defines a convex polytope. In general, a convex polytope is defined as the convex hull of a finite set (of vertices), page 14:
Def: A convex polytope $P\subset \mathbb{R}^n$, or symply a polytope, is defined as the convex hull of a non empty finite set $\{x_1,\dots,x_q\}$
So maybe you should provide the definition of convex polytopes that you are using. There another way to characterize a convex polytope $P$. Which is by means of a finite set of inequalities. Some times this is refered as the H-V theorem were H stands for half-spaces and V for vertices. Sometimes also as the Fundamental Theorem of Polytopes:
Theorem 9.2: A non-empty set $P$ of $\mathbb{R}^n$ is a (convex) polytope if, and only if, it is a bounded polyhedral set.
Polyhedral set is defined in the same text as
Def: A subset $Q$ of $\mathbb{R}^n$ is called a polyhedral set if $Q$ is the intersection of a finite number of closed halfspaces, or $Q=\mathbb{R}^n$.
Best Answer
This isn't true. Take $A = Conv \{0, (1,0) \}$ and $B = Conv \{0, (0,1) \}$. Then $A+B$ is the unit square, and $P = Conv \{0, (1,1)\} \subset A+B$, but there is no $Q \subset A, R \subset B$ such that $P = Q + R$.