@FedorPetrov, my approach to numerical tests is to calculate $S(m, q)(d)^2 - S(m, q)(d-1)S(m, q)(d+1)$ as a polynomial in $q$, bound its largest real root, and test all integers from $2$ up to that bound. Using that approach I've found no counterexamples for $2 \le m \le 8$, $1 \le d < 100$. As a bonus observation, for $m > 4$ the largest upper bound I've found was about $4.4$, so not many values of $q$ need to be tested.

I'm curious about your experimental data. I'm experimenting with $k=m=2$, $f_i$ up to cubic terms, coefficients from $\{-1,0,1\}$ except fixed constant coefficients $-5$ and $-13$ (as a compromise because working over $\mathbb{Q}[a,b]$ and using constant coefficients $-a$ and $-b$ ran into memory trouble), and in hundreds of thousands of tests I haven't yet found a case which doesn't appear to have a linear recurrence of order at most $9$.

Don't you need to specify a term ordering for Gröbner base calculations in order to define $a(n)$? Otherwise if e.g. $f_1 = x_1 - x_2 - 3$ it's arbitrary whether $t(1) = x_1 + x_2 = 2x_1 - 3 = 2x_2 + 3$ yields $a(1) = 0$ or $a(1) = 2$.

Incidentally the "injective" and "such that their images are disjoint", which capture the original definition of $W$ as a "subset of $K$", are unused and unnecessary in this case. If $i \neq j$ then clearly $w(i) \neq w(j)$ and $w(i) \neq \overline{w}(j)$ since otherwise respectively $w(i) \overline{w}(j)$ or $w(i)w(j)$ would be zero, violating the condition. We can weaken to $w(i) \neq \overline{w}(i)$ (in the original formulation, $W[i] \neq W[i+k]$), but that constraint is then only relevant if there are non-zero square roots of zero in $K$.

@joro, I don't understand the question, but it's possible that the edit I just completed already answers it.

1 isn't a zero divisor. That quibble apart, see mathoverflow.net/questions/42647/… for examples.

Isn't this just your previous question rephrased?

@DenisIvanov, there are a lot of theorems about polynomial roots and I don't know which you thought was relevant.

@DenisIvanov, apparently not. The first counterexample is $P_{50539}$.

@DenisIvanov, the easy statements are that powers of 2 are irrelevant (because they certainly don't cause carries); the odd part must be squarefree; and if odd primes $p$ and $q$ both occur then $(p-1) + (q-1)$ must also be without carries.

Yes, precisely.