The coming below is nothing else but thinking loudly.

The differential operator
$$
D=\operatorname{id}+x\frac d{dx}\colon h\mapsto (xh)'
$$
"kills" the unwanted odd powers modulo 2. Indeed, if
$$
h=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+\dots,
$$
then
$$
Dh=\sum_ {n=0}^\infty (n+1)a_nx^n
\equiv\sum_ {k=0}^\infty a_ {2k}x^{2k}\pmod 2
$$
where the congruence is applied to all coefficients in the power
series expansions.
Therefore, the OP asks for the congruence
$$
D\biggl(\frac fg\biggr)\overset?\equiv D(f)\pmod 2
$$
to be true, which after multiplication by $g^2$ becomes (modulo 2) the
congruence
$$
D(fg)\overset?\equiv D(f)g^2\equiv D(fg^2)\pmod{2},
$$
equivalently,
$$
\frac{d}{dx}\bigl(xf(x)g(x)\bigr)
\overset?\equiv\frac{d}{dx}\bigl(xf(x)g(x)^2\bigr)\pmod{2}.\qquad\qquad\qquad(*)
$$

The function $f(x)$ can be in a certain sense eliminated from the required formula
by using
$$
g(x)=\sum_{n=0}^\infty x^{n^2}
=\sum_{m=0}^\infty x^{(2m+1)^2}
+\sum_{m=0}^\infty x^{(2m)^2}
=xf(x^8)+g(x^4)
$$
which implies
$$
f(x)=\frac{g(x^{1/8})-g(x^{1/2})}{x^{1/8}}.
$$

In addition, we can use repeatedly
$$
h(x^2)\equiv h(x)^2\pmod{2}.
$$

If I am correct in my derivation, ($ * $) reduces to
$$
\frac{d}{dx}\bigl(x(g-g^4)(g^8-g^{16})\bigr)
\overset?\equiv0\pmod{2}.\qquad\qquad\qquad(**)
$$