d_r in spectral sequence given by zig-zag

Because each page $E_r$ is not just made from entries of the double complex, but from classes that survive $r-1$ stages of taking cohomology. To compute the next differential, you repeatedly solve “cocycle up to boundary” equations in the double complex, and those equations force a zig-zag path.

Suppose the double complex has horizontal differential $d_h$ and vertical differential $d_v$, with total differential

$$D = d_v + d_h$$

up to the usual sign convention. If an element $a_0 \in A^{p,q}$ represents a class on $E_r^{p,q}$, then surviving to the $r$-th page means roughly that $a_0$ is not merely vertically closed, but that its horizontal differential can be killed by moving through the double complex.

Concretely, one tries to solve equations of the form

$$d_v a_0 = 0,$$

then

$$d_h a_0 = - d_v a_1,$$

then

$$d_h a_1 = - d_v a_2,$$

and so on, until

$$d_h a_{r-2} = - d_v a_{r-1}.$$

Visually, this means:

$$a_0 \;\xrightarrow{d_h}\; d_h a_0 = -d_v a_1,$$

so we move horizontally by $d_h$, then “lift back” vertically by finding $a_1$, then move horizontally again, lift vertically again, etc. That alternating horizontal move and vertical lift is the zig-zag.

After $r-1$ such lifts, the $r$-th differential is represented by the final horizontal differential:

$$d_r[a_0] = [d_h a_{r-1}].$$

This lands $r$ columns over and $1-r$ rows down, so for a cohomological spectral sequence,

$$d_r : E_r^{p,q} \to E_r^{p+r,\,q-r+1}.$$

So the reason $d_r$ is given by zig-zags is that an $E_r$-class is a class whose earlier differentials vanish, and “earlier differentials vanish” means each obstruction can be killed by choosing a vertical preimage. Chaining those choices produces the zig-zag, and the final unresolved obstruction is exactly $d_r$.