Why the descent fixes one sign pattern in the four-square identity

The four bilinear forms in Euler's four-square product identity are not the only ones that turn a product of two sums of four squares into a sum of four squares. Euler also recorded the variant with the same first coordinate and

$z_2=x_1{y}_2-x_2{y}_1+x_3{y}_4-x_4{y}_3,z_3=x_1{y}_3-x_2{y}_4-x_3{y}_1+x_4{y}_2,z_4=x_1{y}_4+x_2{y}_3-x_3{y}_2-x_4{y}_1,$

which is the form Dummit's Lemma 1 displays, and the norm of a product of quaternions gives a third,

$z_1=x_1{y}_1-x_2{y}_2-x_3{y}_3-x_4{y}_4,z_2=x_1{y}_2+x_2{y}_1+x_3{y}_4-x_4{y}_3,z_3=x_1{y}_3-x_2{y}_4+x_3{y}_1+x_4{y}_2,z_4=x_1{y}_4+x_2{y}_3-x_3{y}_2+x_4{y}_1,$

which is the route MIT's Lecture 22 takes. All three are polynomial identities in the eight variables, so any of them proves that a product of two sums of four squares is again one.

The descent asks more of the identity than that. In Descent step: a smaller multiple of $p$ is a sum of four squares the second quadruple is congruent to the first coordinatewise modulo $m$, and what the proof needs is that all four output coordinates then become divisible by $m$. Substituting $y_1\equiv x_1$, $y_2\equiv x_2$, $y_3\equiv x_3$, $y_4\equiv x_4$ modulo $m$ and writing $(a,b,c,d)$ for $(x_1,x_2,x_3,x_4)$, the pattern fixed in Euler's four-square product identity gives

$z_1\equiv a^2+b^2+c^2+d^2,z_2\equiv ab-ba-cd+dc=0,z_3\equiv ac+bd-ca-db=0,z_4\equiv ad-bc+cb-da=0(\mathrm{m}\mathrm{o}\mathrm{d}m),$

and the first of these is the multiple $pm$, hence also $0$ modulo $m$. Euler's second pattern behaves the same way: its last three coordinates become $ab-ba+cd-dc$, $ac-bd-ca+db$ and $ad+bc-cb-da$, all identically $0$. The quaternion pattern does not: under the same substitution its coordinates become $a^2-b^2-c^2-d^2$, $2ab$, $2ac$ and $2ad$, and the hypotheses of Descent step: a smaller multiple of $p$ is a sum of four squares — that $m$ divides $a^2+b^2+c^2+d^2$ and that the second quadruple is the centred residue quadruple of the first — do not force $z_1$, $z_2$, $z_3$ or $z_4$ to vanish modulo $m$.

So the choice of signs is load-bearing for the divisibility step written here, and it is displayed rather than summarised for that reason. This says nothing about whether some other argument can descend with the quaternion pattern; a proof organised around quaternion divisibility rather than around congruences between bilinear forms is a different argument with a different bookkeeping, and nothing above bears on it.