In this question, a functional equation is solved for functions with a power series. We find a recursive formula: (copied from the answer by user achille hui)
\begin{align} ( 2^1 - 3 ) a_2 &= 0\\ ( 2^2 - 4 ) a_3 &= 2 a_2 a_2 \iff 0 = 0 \text{ ( i.e. } a_3\;\text{ is arbitrary )}\\ ( 2^3 - 5 ) a_4 &= 3 a_3 a_2 + 2 a_2 a_3\\ ( 2^4 - 6 ) a_5 &= 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4\\ &\;\vdots \end{align}
According to the answer, it turns out that $\sum a_i x^i$ is equal to $k \sin(x/k)$ or $k \sinh(x/k)$, depending on the sign of $a_3$.
How does one see this? A messy inductive proof will probably prove equality, but that is not what I am looking for. What methods/tricks are there to find a closed form for the coefficents $a_i$ from the recursion, and then recognize it as the power series of $\sin$ or $\sinh$?
The logic is the other way around. If handed the recursion, standard generating function techniques will recognize it as an encoding of the differential equation in the other question, $f(2x) = 2f'(x)f(x)$ and further analysis would be based on the differential equation. The problem is to find particular solutions. You could do this by calculating $a_i$'s for the simplest cases, using the recursion, and looking for a pattern (power series with $a_n = \pm pq^n\frac{1}{(2n+1)!}$ will stand out quickly, or use OEIS), but things are easier than that.
It is not hard to "see" $\sin x$ as a solution, and the problem may have been inspired by the formula for $\sin (2x)$. If you can somehow generate this solution, things get better. The inhomogeneity of the equation can be used: if $f(x)$ is a solution, so is $af(bx)$ for suitable $a$ and $b$, since both parameters will alter the ratio of the two sides in some way, and $b$ can be chosen to reverse the effect of $a$. The condition happens to be $ab=1$, so we now have $k \sin (x/k)$ as a family of solutions. Four of the five solution types at the other question can be obtained from that by taking $k$ to be a complex number (it must be pure imaginary to have a real valued solution on $\mathbb{R}$) or letting it approach $0$ or $\infty$. The $k$-deformation $kf(\frac{x}{k})$ leaves $f'(0)$ constant and by looking at solutions approximating a power function $px^q$ near $0$, one sees that there are very few possibilities.
The freezing of $f'(0)$ and the existence of an additional family of solutions, the $A \exp(Bx)$, with $f'(0)=1/2$, are somewhat unusual, and I suspect that if you relax the condition of analyticity near $0$ there could be huge families of solutions that imitate one of the analytic ones at points in a geometric progression $(2^k x_0), k \in \mathbb{Z}$ and interpolate almost arbitrarily in between, since the equation is of the delay-differential type.
I'll offer the following as one approach to explaining why the solutions look as they do. The analytic solutions satisfy a better functional equation
that specializes to the other one when $x=y$. I don't fully understand why to expect the two functional equations to be equivalent for power series, nor could I have predicted the two variable relation without knowing it for the sine function (and therefore the other four sine-derived solutions) and checking that it works for the exponential solution. But once it is there, solutions are seen to correspond to "half" group representations; the equation holds that for real $f$ and $x$, the imaginary part of $g(x)= f'(x) + if(x)$ satisfies $\Im g(x+y)= \Im (g(x)g(y))$, or
I did not yet complete the analysis of this equation, but it is immediate from this way of writing things that if $g(x)$ is a solution, so is $g(kx)$, and that there are solutions $g(x)=Ae^x$ with $A-A^2 \in \mathbb{R}$. If this is all solutions, it means $A$ real or on the famous line $1/2 + it$, with degenerate solutions for $A = 0,1$ and a special solution at $A=1/2$ where the two lines intersect. These are five possiblities, and one thing to check is if they correspond to the five solution types of the original problem.