It's hard to say without knowing what you mean by "inside." I'll assume that you mean $\iota_i:B_1,B_2\hookrightarrow M$ are embedded open balls of the same (finite) dimension, and your question is, "Is it necessary that $\iota_1(B_1)\cap\iota_2(B_2)$ is homeomorphic to an open ball $D^n$?" In that case, the answer is "no, trivially."
For example, take two two-dimensional discs, wrap them both into "C"-shapes, and arrange them so that the intersection has two components. The intersection is homeomorphic to a disjoint union of two balls (and not homeomorphic to a single ball, since all Euclidean balls are connected).
Perhaps you might insist that the intersection be homeomorphic to a finite union of balls, but I do not think that is true either. Embed two disks into $\mathbb{R}^2$, one disk onto $\{(x,y)\ |\ y>\sin(x)\}$ and the other onto $\{(x,y)\ |\ y<-\sin(x)\}$. If I am not mistaken in my formulas, the intersection of these two sets is homeomorphic to a countable disjoint union of open balls.
These constructions are in two dimensions, with codimension zero. You could conceivably do worse by examining fractals or moving to codimension $>0$.
Topological manifolds are incredibly flexible because homeomorphisms are incredibly flexible. All a topological manifold requires is that it be locally homeomorphic to $\mathbb{R}^n$; nothing more. Since our intuition is usually created by working with smooth maps, general continuous maps, even homeomorphisms, can be very counterintuitive indeed.
(For an example of this counterintuitive behavior that is not strictly a counterexample to your question, consider the Alexander horned sphere. It topologically embeds a disk $D^2\hookrightarrow S^3$ such that one component of the $S^3-D^2$ is homeomorphic to $D^3$ and the other is not $1$-connected, hence most emphatically not homeomorphic to $D^2$. In fact, its fundamental group is not even finitely generated!)
(Edited to correct a sign error.)