I would like to prove the following statement, which I'll state initially without context since I believe it to be purely algebraic.
Let $f(x)=x+a_3x^3+a_5x^5+\cdots$ be an odd formal power series (say, over $\Bbb Q$). Then the following are equivalent:
\begin{align} (1)\quad & f(2x)=\frac{2f(x)f'(x)}{1-\varepsilon f(x)^4}\\ \\ (2)\quad & f(u+v)=\frac{f'(u)f(v)+f(u)f'(v)}{1-\varepsilon f(u)^2f(v)^2}\\ \\ (3)\quad & f'(x)^2=1-2\delta f(x)^2+\varepsilon f(x)^4 \end{align}
Here, $\delta=-3a_3$ and $\varepsilon=10a_5-3a_3^2$ (more on this below).
Here was the original setting: suppose we have a genus $\varphi$ with characteristic power series $Q(x)=1+c_2x^2+c_4x^4$. That is, $Q$ is even and has constant term $1$. Then write $f(x)=x/Q(x)$ to obtain an odd power series $f(x)$ starting with $x$.
Via algebraic trickery, one can show that the power series $$h(y)=h(f(x))\colon=\frac{f(2x)}{2f(x)f'(x)}$$ gives the value of the genus $\varphi$ on quaternionic projective spaces $\Bbb H P^n$. More precisely, $h(y)$ is even, and $$h(y)=1+\varphi(\Bbb H P^1)y^2+\varphi(\Bbb H P^2)y^4+\cdots.$$ If $h(y)$ has the particularly nice form $h(y)=\frac{1}{1-\varepsilon y^4}$, we say that $\varphi$ is elliptic, in which case we immediately see $$\varphi(\Bbb H P^n)=\begin{cases}\varepsilon^{n/2},&n\text{ even}\\ 0,&n\text{ odd} \end{cases}.$$ So geometrically, $\varepsilon$ is just the genus of $\Bbb H P^2$. Now writing $y=f(x)$, let $g(y)$ be the formal inverse function (so $g(y)=x$). (It follows that $f'(x)g'(y)=1$.) After some more manipulation, one finds that $g'(y)$ describes the genus of complex projective spaces, i.e., $$g'(y)=1+\varphi(\Bbb C P^2)y^2+\varphi(\Bbb CP^4)y^4+\cdots,$$ and the number $\delta$ is exactly $\varphi(\Bbb CP^2)$.
This story came from chapter one of "Manifolds and Modular Forms" by Hirzebruch, Berger, and Jung. The numbers $\delta$ and $\varepsilon$ are the same constants used to parameterize elliptic curves, as the authors explain in more detail in chapter 2.
Before even mentioning elliptic curves, however, the authors leave as an exercise to prove $(1)\implies(2)$, as above. I can't tell if I am missing some well-known trick, or if these identities are actually difficult to establish. I know very little about elliptic geometry, so I struggled to find help online.
Partial answers are welcome, and hopefully will illuminate how to go about the rest. As of now, I can't prove any implications except the obvious $(2)\implies(1)$. I've made progress on $(1)\implies(2)$: if I knew $f'(x)^2$ could be written $1+\alpha f(x)^2+\beta f(x)^4$ for some constants $\alpha,\beta$, it's fairly straightforward to determine the constants.
Well this got a non-zero amount of attention, so for the first time I'm answering my own question. Apologies for the length. A power series $f(x)\in\Bbb Q[\![x]\!]$ is admissible if $f(x)$ is odd and $f'(0)=1$.
Claim 1: for a fixed $\delta,\varepsilon$, there is a unique admissible power series $f_1(x)$ satisfying condition $(1)$. The analogous statement holds for condition $(3)$ (and of course for $(2)$, but we can avoid proving this directly).
Claim 2: $(3)\implies(2)\implies(1)$
Claim 3: $(1)\implies(3)$.
Proof of claim $3$: let $f_1$ be an admissible power series satisfying $(1)$. There is a unique admissible power series $f_3$ with the same parameter $\varepsilon$ satisfying $(3)$, and by claim $2$, $f_3$ satisfies $(1)$. By uniqueness, $f_1=f_3$, so $(1)\implies(3)$.
Proof of claim 1: to show uniqueness, by claim $2$, it suffices to show uniqueness for condition (1). We have that $f(x)f'(x)=\frac12f(2x)(1-\varepsilon f(x)^4)$. Write out $f(x)=x+a_3x^3+a_5x^5+\cdots$. Compare coefficients of $x^{2k+1}$ to see $$(2k+2)a_{2k+1}+\{\text{expression in }a_3,\ldots, a_{2k-1}\}=2^{2k}a_{2k+1}+\{\text{expression in }a_3,\ldots, a_{2k-1}\}.$$
It follows that $a_{2k+1}$ can be written in terms of $a_3,\ldots, a_{2k-1}$ and $\varepsilon$ provided that $(2k+2)\neq 2^{2k}$. The only issue is when $k=1$, but $a_3$ is already determined by $\delta$.
This also establishes existence for condition $(1)$. To show existence for condition $(3)$, we use basic calculus. Since we can write $f'(x)=\sqrt{1-2\delta f(x)^2+\varepsilon f(x)^4}$, the chain rule shows that $$f^{-1}(x)=\int_0^x\frac{dt}{\sqrt{1-2\delta t^2+\varepsilon t^4}}.$$ Recall that a formal power series $f$ has an inverse if $f(0)=0$ and $f'(0)=1$, so the previous statement makes sense and determines $f$.
Proof of claim 2: $(2)\implies(1)$ is obvious, so the bulk of the work is showing $(3)\implies(2)$. To do so, we again use single variable calculus. Note that it is enough to prove equality on an open interval, so this entails no loss of generality. Make the following definitions: \begin{align*} u&=f(x)\\ v&=f(y)\\ U&=f'(x)=\sqrt{1-2\delta u^2+\varepsilon u^4}\\ V&=f'(y)=\sqrt{1-2\delta v^2+\varepsilon v^4}\\ r&=\frac{uV+Uv}{1-\varepsilon u^2v^2} \end{align*}
Using basic calculus, observe that $g(f(x))=x$ implies $g'(f(x))f'(x)=1$, so $g'(y)=1/f'(x)$. Then with a slight abuse of notation, $$\int_0^k \frac{1}{f'(x)}dy=\int_0^kg'(y)dy=g(k)-g(0)=g(k). $$
If, in addition, $f'(x)$ can be written in terms of $y=f(x)$, then we can recover the inverse function $g$ through this process. For example, if $y=\sin x$, then $y'=\cos x=\sqrt{1-\sin^2 x}=\sqrt{1-y^2}$, and thus $\int_0^k\frac{1}{\sqrt{1-y^2}}dy=\arcsin k$ (at least on some open interval of values for $x,y$, and $k$). In our case, this means $\int_0^u\frac{du}{U}=g(u)=x$. We therefore will prove $\frac{du}{U}+\frac{dv}{V}=0$. We will consider $v$ as a function of $u$ in such a way that $r$ is a constant. Note that $u=0$ gives $v=r$, so we can integrate $\frac{du}{U}+\frac{dv}{V}=0$ and use a change of variables to see
$$\int_0^u\frac{du}{U}+\int_r^v\frac{dv}{V}=0 $$
which implies
$$\int_0^u\frac{du}{U}+\int_0^v\frac{dv}{V}=\int_0^r\frac{dv}{V}.$$
From our previous comments, this means $g(u)+g(v)=g(r)$, and thus $x+y=g(r)$, or $f(x+y)=r$. This is exactly the statement of $(3)$, which establishes our reduction.
To finish the proof of $(3)\implies(2)$, it remains to see $\frac{du}{U}+\frac{dv}{V}=0$. To this end, we now regard $v$ as a function of $u$ which makes $r$ constant. It suffices to let $u$ take values in some narrow open interval of positive real numbers, and we thus avoid the issue of whether $U$ and $V$ are defined. Since $r$ is constant, we differentiate both sides of $r=\frac{uV+Uv}{1-\varepsilon u^2v^2}$ with respect to $u$ to find
$$0=\frac{\left(V+\frac{2\varepsilon uv^3-2\delta uv}{V}\frac{dv}{du}+\frac{2\varepsilon u^3v-2\delta uv}{U}+U\frac{dv}{du}\right)(1-\varepsilon u^2v^2)+(uV+Uv)(2\varepsilon uv^2+2\varepsilon u^2v)\frac{dv}{du}}{(1-\varepsilon u^2v^2)^2}.$$ Substituting in $r$, simplifying, and rearranging gives $$\frac{du}{U}\left(UV+2\varepsilon u^3v-2\delta uv+2\varepsilon uv^2rU\right)+\frac{dv}{V}\left(UV+2\varepsilon uv^3-2\delta uv+2\varepsilon u^2vrV\right)=0.$$ The expressions in parenthesis are both nonzero for small values of $u,v$, so it now suffices to show they are equal. After cancelling, it suffices to show $$u^2+vrU=v^2+urV.$$
We compute \begin{align*} u^2+vrU-v^2-urV&=u^2+\frac{uvUV+U^2v^2}{1-\varepsilon u^2v^2}-v^2-\frac{uvUV+u^2V^2}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{U^2v^2-u^2V^2}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{(1-2\delta u^2+\varepsilon u^4)v^2-u^2(1-2\delta v^2+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{(1+\varepsilon u^4)v^2-u^2(1+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=\frac{u^2(1-\varepsilon u^2v^2)-v^2(1-\varepsilon u^2v^2)+v^2(1+\varepsilon u^4)-u^2(1+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=0, \end{align*} which establishes $(3)\implies (2)$.