Construct a new boundary operator for each Dp into Dp-1, based on the singular boundary operator. This is the usual machinery; a simplex at dimension p becomes an alternating sum of its faces at dimension p-1. But does this make any sense? We are only considering chains (i.e. linear combinations of simplexes) in the relative homology group, hence the boundary becomes a linear combination of simplexes that sum to 0 in the quotient group, or if you prefer, the boundary becomes a linear combination of simplexes that lie in Sp-1. So indeed this could be an element in Dp-1. Let's examine this in detail, focusing on D5 for illustrative purposes.
Let the chain y determine a member of h5(S5/S4). For notational convenience, let A = S5, B = S4, C = S3, and A5 is the group of 5-chains in A, C4 the group of 4-chains in C, etc. The homology group h5(A/B) is based on chains of simplexes in A whose boundaries lie in B. We then mod out by those chains that already lie in B, and by the image from above, which is the boundary of A6. Thus y represents a coset of bound(A6)+B5. Since bound2(y) = 0, bound(y) is automatically a chain in any space it lives in, in this case B4. Adjust y by some z in bound(A6) or B5. Does the boundary of y still become the same element in D4? If z is in A6, then boundary squared is 0, so that is not a problem. If z is in B5, its boundary is part of the image from above, and becomes 0 in the relative homology h4(B/C). Therefore the map from D5 to D4 is well defined.
Since singular boundary respects addition, The map from Dp to Dp-1 is a group homomorphism. Furthermore, since boundary squared is 0, the map from Dp to Dp-1 to D p-2 = 0. Therefore the collection of homomorphisms down the skeletons of S forms a chain complex, and each homomorphism is thus called a cellular boundary operator. Naturally this leads to homology, but before we go there, let's look at the cellular boundary operator in more detail.
Continue the A B C notation above, so that C is a subspace of B, is a subspace of A. Write the long exact sequence for this triple homology as follows.
… h5(B/C) → h5(A/C) → h5(A/B) → h4(B/C) …
The last function, from h5 to h4, looks like the cellular boundary, but is it? As you recall, the long exact sequence is based on the serpent lemma, which is based on a commutative diagram. In this case the down arrows are the simplex boundary operators, so the function is based on the simplex boundary. This is in fact how the cellular boundary operator is defined, so everybody's happy.
Can we get more specific about the simplexes that generate these homology groups? Remember that a simplex, mapped homeomorphically onto a ball, generates the homology of that ball relative to its boundary. Also, the characteristic function f from a disjoint union of balls into S induces a relative homology isomorphism. Therefore we can apply f, and let the closed p cells of S generate Dp. The cellular boundary operator produces the boundary of the p cell, which must be viewed as a chain in Sp-1/Sp-2.