First of all, Wikipedia article is not meant to be a detailed technical explanation of how the proof of PC goes, only to give a casual reader with limited math background a general idea.
Here is what "caps" in Perelman's proof are. Suppose that $M$ is a simply-connected closed (compact with empty boundary) smooth 3-dimensional manifold and $S\subset M$ a smoothly embedded 2-dimensional sphere. Since $M$ is simply-connected, $S$ separates $M$ in two components, so removing it from $M$ yields two noncompact manifolds $N_1, N_2$. Both $N_1, N_2$ admit natural compactifications as smooth compact manifolds $\bar{N}_i, i=1,2$, with spherical boundaries. Attach (smoothly) 3-dimensional balls $B_1, B_2$ to $\bar{N}_1$, $\bar{N}_2$ along the boundary spheres. This results in closed 3-dimensional manifolds $M_1, M_2$. The "cups" are these 3-dimensional balls $B_1, B_2$. This process of decomposition of $M$ is called "connected sum decomposition":
$$
M= M_1\# M_2.
$$
It is vaguely similar to the process of decomposition of a natural number as a product of two smaller natural numbers, say
$$
12=6\cdot 2,
$$
or, more appropriately,
$$
1=1\cdot 1.
$$
Repeating this process with $M_1, M_2$ inductively finitely many times, one obtains a connected sum decomposition
$$
M= L_1\# L_2 \# ... \#L_n.
$$
The way Perelman's proof works is that after finitely many decompositions, each $L_i$ is shown to be diffeomorphic to the usual 3-dimensional sphere $S^3$ (and this is the heart of the proof). Therefore,
$$
M= S^3\# ... \# S^3.
$$
It is a standard (and easy) fact of geometric topology that if $K$ is any (smooth) connected manifold of dimension $m$ then
$$
K\# S^m=K,
$$
where $K$ means "diffeomorphic." In analogy with the product decomposition of natural numbers, $S^m$ plays the role of the unit: $$k\cdot 1=1$$
for all natural numbers $k$. (In fact, the analogy is deeper: The connected sum operation is commutative and associative. Moreover, every compact oriented connected 3-dimensional manifold $M$ admits a "prime decomposition" as a connected sum of manifolds which are not decomposable any further, and these "prime factors" are unique.)
In the setting of Perelman's proof, one concludes that
$$
M= S^3\# ... \# S^3= S^3,
$$
thereby proving the PC.
