I can now, with less computer calculation than I'd feared, answer Question 1. (I'll say more about Question 2 later).

Lemma:____ Let V be the vector space over Z/2 spanned by the C(r1,r2,r3) and the C(s1,s2,s3,s4). If an element of V has its power series expansion divisible by x^(l^2), it is 0.

To see this, let K be an algebraic closure of Z/2, S' be the subring of K[[x]] generated over K by the [j] and L be the field of fractions of S'. I can show that L/K is the function field of a curve, that there are exactly l(l-1)(l+1)/24 valuation rings in L/K that don't contain S', and that each of [1],...,[l-1] has a simple pole at each of them. So an element of V has at most l(l-1)(l+1)/6 poles in L/K, counted with multiplicity.

Also, the localization of S' at the maximal ideal generated by [1],...,[l-1] is dominated by exactly (l-1)/2 valuation rings in L/K. I can show that for each r prime to l there is an automorphism of L/K taking [j] to [rj] for each j, and that these automorphisms act transitively on this set of valuation rings. Now the elements of V are fixed by these automorphisms. So an element of V whose power series expansion is divisible by x^(l^2) has zeros of order at least l^2 at each of these valuation rings. Since (l^2)(l-1)/2 > l(l-1)(l+1)/6, such an element must vanish.

Suppose now that l=23. Since l=7 mod 16, my answer to my question "Existence of certain identities..." shows that C is in V. The lemma then shows that to prove 2. it's enough to show that C(3,3,1,2)+C(1,3,6)+C is divisible by x^529 in Z/2[[x]]. Now C(3,3,1,2)+C(1,3,6) is a sum of monomials in the [j]. Replacing each [j] by the sum of the first 2 terms in its power series expansion only modifies the sum by something divisible by x^529, and we're reduced to an easy computer calculation.

Establishing 3. and 4. is harder since we don't know in advance that C is in V. I'll use:

Theorem____Suppose there is an element R of V such that R+C is divisible by x^(d+1) where d=l(l+1)(l+1). Then R=C.

To see this, recall that F=x+x^9+x^25+..., that G=F(x^l), that H=G(x^l) and that C^2+C=G+H. Now there is a symmetric degree l+1 2-variable polynomial P over Z/2 with P(F,G)=0; furthermore P(z,H) is monic in z of degree l+1. This is discussed in my question "What's known about the mod 2 reduction...". Suppose a^l=1. Replacing x by ax^l in the identity P(F,G)=0, we get a root of P(z,H)=0 of the form ax^l+... This gives us l distinct roots, with G among them. By symmetry P(H,G)=0. So H(x^l) is still another root, and P(z,H) factors into linear factors over K[[x]].

Now R^2+R+G+H=(R+C)^2+ (R+C), and so is divisible by x^(d+1). Since P(G,H)=0, P(R^2+R+H,H) is divisible by x^(d+1). Also R^2+R has poles of order at most 8 at each valuation ring in L/K that doesn't contain S', while H has poles of order at most 12. It follows that P(R^2+R+H,H) has at most (l+1)l(l-1)(l+1)/2 =d(l-1)/2 poles, counted with multiplicity, in L/K. Arguing as in the proof of the lemma we see that it has more than d(l-1)/2 zeros, counted with multiplicity. So it vanishes, and R^2+R+H is a root of P(z,H)=0. Examining the roots of this equation we see that R^2+R+H can only be G. So R^2+R=G+H, and R=C.

Suppose now that l=31. To prove 3. we now see that it suffices to show that C(3,3,2,3)+C(2,3,7)+C is divisible by x^(186^2), since 186^2 >(31)(32)(32). This is carried out as in the case l=23, but now we have to use the first 12 terms in the power series expansion of each [j] rather than just the first 2. The treatment of 4. is similar, but now we show divisibility by x^(329^2) using the first 14 terms in the power series expansion of each [j].