chain and boundary
To integrate a form you need something to integrate over, and a single smooth patch is rarely enough — a real region might be several pieces glued together, each with its own orientation. A chain is the bookkeeping device for this: a formal sum of oriented k-dimensional pieces (called k-cells or simplices), with integer coefficients that let you count a piece twice, or with a minus sign to flip its orientation. A 1-chain is a sum of oriented curve segments; a 2-chain is a patchwork of oriented surface pieces.
The boundary operator, written with a curly d (the symbol used here as 'bd'), takes a k-chain and returns its (k-1)-dimensional boundary as a (k-1)-chain — and it tracks orientation carefully. The boundary of an oriented curve segment is its endpoint minus its start point. The boundary of an oriented disk is its rim, traversed counterclockwise. The boundary of a solid ball is its sphere. The signs are arranged so that when two cells share an edge, that shared edge appears with opposite orientations from the two sides and cancels in the sum — which is exactly why interior cuts disappear when you glue patches together.
The decisive fact, the mirror image of d squared equals zero, is that the boundary of a boundary is empty: bd(bd of c) = 0. A surface has an edge, but that edge is a closed loop with no endpoints of its own; a solid has a surface, but that surface is closed with no rim. This is the geometric twin of d squared = 0, and the two together are what make the generalized Stokes' theorem (integral of d omega over a region equals integral of omega over its boundary) a perfect duality between the operator d on forms and the operator boundary on chains.
Let c be the oriented straight segment from A to B. Its boundary is bd c = B - A (the endpoint counted +1, the start counted -1). For the 1-form df, the integral over c is f(B) - f(A) — and you can read this as the integral of f over bd c, the boundary, which is Stokes' theorem at the lowest degree.
The boundary of a segment is its endpoints with signs; this is the bookkeeping that powers Stokes' theorem.
A chain is a formal sum, not a set: the coefficient -1 does not mean 'remove' but 'include with reversed orientation', and 2 times a cell is a legitimate chain. This algebraic bookkeeping, not a geometric picture, is what makes boundaries cancel cleanly.