From L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXII, page $644$
$$ A^4+B^4=C^4+D^4. $$ L. Euler$^{165}$ took
$\,A=p+q,\;\;D=p-q,\;\;C=r+s,\;\;B=r-s\,$
and derived
$$(1)\qquad\qquad\qquad pq(p^2+q^2)=rs(r^2+s^2).$$
Set $\,p=ax,\,q=by,\,r=kx,\,s=y.\,$ Then
$$ y^2/x^2=(k^3-a^3b)/(ab^3-k). $$
If $\,k=ab,\,x=1,\,$ then $\,y=\pm a,\,C=\pm A,\,B=\mp D.\,$
Set therefore $\,k=ab(1+z).\,$
Then $\, y^2/x^2=a^2Q/(b^2-1-z)^2,\,$ where
$$ Q\!=\!(b^2\!-\!1)^2\!+\!(b^2\!-\!1)(3b^2-1)z \!+\!
3b^2(b^2\!-\!2)z^2\!+\!b^2(b^2\!-\!4)z^3\!-\!b^2z^4. $$
Let $\,Q\,$ be the square of $\,b^2-1+fz+gz^2\,$ and
choose $\,f,g\,$ to make the terms in $\,z,z^2\,$ agree. Thus,
$$ f=\frac{3b^2-1}2,\quad g=\frac{3b^4-18b^2-1}{8(b^2-1)},\quad
z=\frac{b^2(b^2-4)-2fg}{b^2+g^2}. $$
Then $\,x:y=b^2-1-z:a(b^2-1+fz+gz^2).\,$
As examples, Euler took $\,b=2,\,b=3\,$ and found the solution
$$ A=2219449,\quad B=-555617,\quad C=1584749,\quad D=2061283,$$
and an erroneous$^{166}$ one replaced in his next paper by
$$ A=12231,\quad B=2903,\quad C=10381,\quad D=10203.$$
I have verified the steps for Euler as quoted in Dickson. For $\,b=1\,$
the result is $\,A=1,\,B=-1,\,C=1,\,D=-1\,$ and for $\,b=2,\,b=3\,$ the
results are as given.
The general result given by Euler depends on one parameter $\,b.\,$
Define the two variable polynomial
$$ P(u,v) := u^{13}+27 u^{12} v-214 u^{11} v^2-186 u^{10} v^3-
2481 u^9 v^4+861 u^8 v^5-2804 u^7 v^6\\-972 u^6 v^7-2481 u^5 v^8-
27 u^4 v^9-214 u^3 v^{10}+294 u^2 v^{11}+u v^{12}+3 v^{13}. $$
The solution given by Euler is essentially
$$ A=P(-b,1),\quad B=P(1,b),\quad C=-P(1,-b),\quad D=P(b,1).$$
Note that $\,P(u,v) = {}^h\!f_2(u+v,v-u)\,$ is the connection with
your solution $\,f_2.$
The polynomial corresponding to your $\,f_1\,$ is
$$ P(u,v) := u^6 v+3 u^5 v^2-2 u^4 v^3+u^2 v^5+v^7 $$
due to Gerardin which I got from A Collection of Algebraic Identities by Tito Piezas III.
Dickson mentions Gerardin briefly on page 647
A. Gerardin$^{181a}$ noted that $(1)$ has the solution
$$ p=a^7+a^5-2a^3+a,\quad q=3a^2,\quad r=a^6-2a^4+a^2+1,\quad s=3a^5,$$
which is simpler than Euler's solution $(5)$.
Your question
How does one find such formulae?
is a difficult one. Another answer mentions
$K3$ surfaces
of which I know almost nothing. Particular cases of interest
include $\,A^4+B^4+C^4+D^4=0\,$ which is the Fermat quartic.
The article
Euler's sum of powers conjecture
mentions the case $\,A^4+B^4+C^4=D^4\,$ which was solved by
Noam Elkies using elliptic curves.