The following questions arise from modular form theory. But this theory isn't needed to formulate or understand them, and I'm not using the modular-forms tag.

NOTATION

Fix an odd prime $N$. Let 
$$
F=\sum_{n\text{ an odd square}}x^n\in\mathbb{Z}/2[[x]].
$$
Set $G=F(x^N)$. There is a degree $N+1$ irreducible polynomial relation between $F$ and $G$ over $\mathbb{Z}/2$, (The relation has the form $(F+G)^{N+1}+(\text{lower degree terms}) =0$, and is symmetric).

Examples: Let $S=F+G$ and $P=FG$. When $N=3$, $S^4=P$. When $N=5$, $S^6=P$. When $N=7$, $S^8=P^2+P$. When $N=11$, $(S^4+P)^3=P$.

Now let $M$ be the integral closure of $\mathbb{Z}/2[G]$ in the degree $N+1$ extension field of $\mathbb{Z}/2(G)$ generated by $F$. View $M$ as a subring of $\mathbb{Z}/2[[x]]$, and let $M(\text{odd})$ consist of those $g$ in $M$ for which each exponent $n$ appearing in $g$ is odd. The trace map $\mathbb{Z}/2(F,G)\to \mathbb{Z}/2(G)$ maps the $\mathbb{Z}/2[G^2]$ module $M(\text{odd})$ into the cyclic module generated by $G$. Let $M_0$ consist of all elements of $M(\text{odd})$ of trace 0. $M_0$ is a free rank $N$ module over $\mathbb{Z}/2[G^2]$.

SOME REMARKABLE FACTS

For small $N$ there are very nice bases of $M(\text{odd})$ and $M_0$ over $\mathbb{Z}/2[G^2]$.

$N=3$... Let $C_1=F$, $C_3=G$, $C_5=(F^2)(G)$. Then the $C_j$ are a basis of $M_0$. There is an element of $M(\text{odd})$ whose trace is $G$; it follows that this element and the $C_j$ form a basis of $M(\text{odd})$. Furthermore for each exponent $n$ appearing in $C_j$, $n\equiv j \mod{8}$ and the Legendre symbol $(n/3)$ is either $(j/3)$ or 0.

$N=5$... Let $C_1=F$, $C_3=(F^3)+(G^2)(F)$, $C_5=G$, $C_7=(F^2)(G)$, $C_9=(F^4)(G)$. Then the above results continue to hold, with 3 replaced by 5 in the final sentence.

When $N=7$ or 11 one can write down $C_j$ where $j$ is odd and $<2N$, and prove the corresponding facts. But when $N=11$ it's not possible for all the $C_j$ to be in $Z/2[F,G]$; nevertheless one can arrange that the product of each $C_j$ by $(1+G^8)$ is in $Z/2[F,G]$.

QUESTIONS

To what extent do the above results generalize to larger $N$? More precisely, let $C(\text{plus})$, (resp. $C(\text{minus})$), consist of those $g$ in $M(\text{odd})$ in which each exponent $n$ that appears has $(n/N)$ equal to 0 or 1 (resp. -1). Let $C$ be $C(\text{plus})+C(\text{minus})$.

Question 1--- Does $M_0=C$?

Remark 1... $M$ is indeed the Serre Swinnerton-Dyer ring of characteristic 2 modular power series for $\Gamma_0(N)$, though this isn't obvious. This allows one to introduce Hecke
operators.

Remark 2... By using these operators one can show that $C$ has rank $N$ or $N+1$. The bases that I've exhibited when $N=3,5,7$ or 11 show that in each of these cases $M_0$ is contained in $C$. When $N=3,5$ or 7, I can show that $M_0=C$. This is surely also true when $N=11$, though I haven't proved it.

Remark 3... For each odd $n$ prime to $N$ there is a formal Hecke operator $\mathbb{Z}/2[[x]]\to\mathbb{Z}/2[[x]]$. As I've indicated, Remark 1 can be used to show that the $T_n$ stabilize $M$, $M(\text{odd})$ and $C$. So one may ask the perhaps more accessible weakening of Question 1:

Question 2... Do the $T_n$ stabilize $M_0$?

EDIT... I'll present precise results for N=3,5, and 7, saving those for N=11 for later. In each case I'll give a Z/2 basis and a "naive" Z/2[G^2] basis of M(odd) and the traces of the naive basis elements. Then I'll write each C_j as a combination of the naive basis elements.   
___But I won't show that the exponents appearing in C_j have the desired properties--this is hard in some cases. For N=3 and 5 my C_j will differ slightly from the ones I gave above but
will satisfy the same conditions that I imposed. Let F,G,S, and P be as above. Instead of using modular forms, I'll use the explicit modular equations connecting S and P.

N=3___ Let G=R/S. Since S^4=P, S^2=(G/S)(F/S)=R^2+R. I'll write this as S^2=(1,2), and adopt  a similar shorthand for any sum of powers of R. The(k)S are seen by induction to lie in Z/2[F,G]; they form a Z/2 basis of M(odd). Since G=(1)S, (G^2)S=(3,4)S. It follows easily that (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis. Their traces are 0,0,G, and G. The C_j are:

C1=F =(0,1)S

C3=G =(1)S

C5=(S^2)G=(2,3)S


N=5___Let R=S^2+(G/S). Then R^2+R=S^2+S^4+(FG/S^2)=S^2. So S^2=(1,2), and once again the (k)S lie in Z/2[F,G] and are a Z/2 basis of M(odd). Since G=RS+S^3=(1)S+(1,2)S=(2)S, it follows that (G^2)S=(5,6)S. Then (0)S, (1)S, (2)S, (3)S, (4)S, and (5)S are a Z/2[G^2] basis. The traces of these elements are 0,0,0,G,G, and G. The C_j are:

C1=F=(0,1)S

C3=(S^2)F=(1,2,3,4)S

C5=G=(2)S

C7=(S^2)G=(3,4)S

C9=(S^4)G+G^3=(4,5)S


N=7___Let R=S^2+S^4+P. Then R^2+R=S^2+S^8+P^2+P=S^2 once again, and R^2=S^4+S^8+P^2=S^4+P. Set J=G+RS. Then J lies in M(odd), and we find that the (k)J and the (k)S are a Z/2 basis of M(odd). 

___Now J^2+JS=(G+RS)(G+(R+1)S)=G(G+S)+S^4=P+S^4=(2). Also since G=(0)J+(1)S, G^2=(0)(J^2)+(2)(S^2). Combining these equations with those of the last paragraph we find that:

(*)__(G^2)J=((1,3,4)J+(2)S, and (G^2)S=(1,2)J+(2,3,4)S.

---From (*) we deduce that (0)J, (1)J, (2)J, (3)J, (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis of M(odd). Their traces are 0,0,G,0,0,0,0, and G. The C_j are given by:

C1=F=(0)J+(0,1)S

C3=(0,1)J

C5=(1)J

C7=G=(0)J+(1)S

C9=(S^2)G=(1,2)J+(2,3)S

C11=(S^4)*G=(2)J+(3)S+(G^2)C_1

C13=(5)J=(1,2,3)J+(2,3)S+(G^2)C_3