When I ask a question that vexes me on MO it sometimes adds enough clarity and incentive that I'm able too find the answer--that's been the case here. In particular the "variation on a theme" in my question also has a proof, along the same same lines as the answer I provided to the original question, which I'll now sketch.

The main actors are now two "Hecke operators" T_5 and U_5 on Z/2[[x]] and three elements of this ring that I'll again call F,G, and r. T_5 and U_5 are defined exactly as were T_3 and U_3 in my first answer, but with 3 replaced by 5. F is the same, G is F(x^5), and r has the same definition with {1,2,3,6} replaced by {1,2,5,10}.

One may show that (F+G)^6=FG,and that F and G are r^6+r^5+r^2+r and r^6+r^5. Also T_5 takes F,F^2,F^3, F^4 and F^5 to 0,0,0,0 and F; furthermore T_5(F^n) is a sum of F^k with k< n and congruent to n mod 2. S is now the map taking f to f+U_5(f).

Lemma 1-- Let g be a sum of (F^i)(G^j )with each i at most 4 and each i+j< d and odd. Then S^2(g) is a sum of F^k with k< d and odd.

We may assume that g=(F^i)(G^j) with i+j=d. Then since i is at most 4, T_5 kills F^i, and U_5(g)=(F^j)(G^i). Then (U_5)^2 takes g to (F^i)(G^j+T_5(F^j)). So S^2 takes g to (F^i)(T_5(F^j)), and we use the fact mentioned above.

Lemma 2-- G=(r^4)(r^2+r), F=(r^4+1)(r^2+r), (F+G)^3=(r^4+r^2)(r^2+r), (F^2+G^2)G=(r^8+r^6)(r^2+r), and (F^4+G^4+FG+G^2)G=(r^10+r^8)(r^2+r).

This follows easily from the expressions of F and G in terms of r.

Lemma 3-- Fix m and let V be the space of dimension 5m+5 spanned by the (u)(G^2n) with n at most m and u in {G,F,(F+G)^3,(F^2+G^2)G,(F^4+G^4+FG+G^2)G}. Then, S^2 maps V into the space of dimension m+2 spanned by the F^k with k< 2m+5 and odd. Furthermore V itself is contained in the space V * of elements (g(r^2))(r^2+r) with g a polynomial of degree at most 6m+5.

The first result follows from Lemma 1. Since G^2n=(r^12+r^10)^n, the second result follows from Lemma 2.

By Lemma 3, the kernel of S^2 restricted to V has dimension at least(5m+5)-(m+2)= 4m+3. So the same holds for the kernel of S^2 restricted to V *. Also it's not hard to show that S stabilizes V *. So the kernel of S: V * --> V * has dimension at least 2m+2.

We now proceed as in my first answer, letting W consist of the elements of Z/2[x] of degree at most 6m+5. The map g-->(g(r^2))(r^2+r) identifies W with V* and we verify that under this identification S: W-->W is the map taking x^n to c(n) where c(n) is defined by the recursion and the initial conditions of the "variation on a theme". Now an induction shows that c(n) has degree n-1 if n is 1,3,4 or 5 mod 6, and degree < n-1 if n is 0 or 2 mod 6. So the kernel of S on W has a basis of elements g_i of degree d_i where each i is an integer in [0,6m+5] that is 0 or 2 mod 6. Since the kernel has dimension at least 2m+2, there is an element of the kernel corresponding to every one of these 2m+2 integers, and we're done.

EDIT (12/7/16)

This answer deals with the "VARIATIONS ON A THEME". Again my earlier answer was sketchy, and I'm replacing it by references to my arXiv preprint (1612.01599 [math.NT], "A characteristic 2 recurrence attached to U_5 with a Hecke algebra application"). The entirely elementary (but seemingly mysterious) argument takes place in the few pages between Definition 1.1 and Theorem 2.11 of that note. Looking at the material preceding Definition 1.1 will help in understanding the mysteries.

The rest of the note uses Theorem 2.11 (slightly strengthened) to study the Hecke operators T_p acting on a space, K, consisting of the mod 2 modular forms of level Gamma_0 (5) annihilated by U_5+I. It's shown that each T_p, p>5, is (uniquely) in its action on K a power series with zero constant term in T_3 and T_7.

Lemma 2-- G=(r^4)(r^2+r), F=(r^4+1)(r^2+r), (F+G)^3=(r^4+r^2)(r^2+r), (F^2+G^2)G=(r^8+r^6)(r^2+r), and (F^4+G^4+F^G+G^2)G=(r^10+r^8)(r^2+r).

Lemma 3-- Fix m and let V be the space of dimension 5m+5 spanned by the (u)(G^2n) with n at most m and u in {G,F,(F+G)^3,(F^2+G^2)G,(F^4+G^4+F*G+G^2)G}. Then, S^2 maps V into the space of dimension m+2 spanned by the F^k with k< 2m+5 and odd. Furthermore V itself is contained in the space V * of elements (g(r^2))(r^2+r) with g a polynomial of degree at most 6m+5.

Lemma 2-- G=(r^4)(r^2+r), F=(r^4+1)(r^2+r), (F+G)^3=(r^4+r^2)(r^2+r), (F^2+G^2)G=(r^8+r^6)(r^2+r), and (F^4+G^4+FG+G^2)G=(r^10+r^8)(r^2+r).

Lemma 3-- Fix m and let V be the space of dimension 5m+5 spanned by the (u)(G^2n) with n at most m and u in {G,F,(F+G)^3,(F^2+G^2)G,(F^4+G^4+FG+G^2)G}. Then, S^2 maps V into the space of dimension m+2 spanned by the F^k with k< 2m+5 and odd. Furthermore V itself is contained in the space V * of elements (g(r^2))(r^2+r) with g a polynomial of degree at most 6m+5.