From where do we get $\varphi(w) = w + \frac 1 w$ for $w^2$ and $z^2 - 2$? In Iteration of Quadratic Polynomials, Julia Sets,

we must find some (invertible?) $\varphi$ s.t. $g = \varphi^{-1}(f(\varphi(z)))$ in order to study $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ the iterates of $g$ assuming we have studied $\{f^{\circ n}(z)\}_{n=1}^{\infty}$ the iterates of $f$


Case 1.
For $f(z) = z^2 + c$ and $g(z) = az^2+bz+d$
we can study $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ by studying $\{f^{\circ n}(z)\}_{n=1}^{\infty}$, the iterates of $f$ because $$g^{\circ n}(z) = \varphi_1^{-1}(f^{\circ n}(\varphi_1(z)))$$
where $$\varphi_1(z) = az + \frac b 2$$ for some appropriate domain and range
Case 2.
For $g(w) = w^2$ and $f(z) = z^2 - 2$
we can study $\{f^{\circ n}(z)\}_{n=1}^{\infty}$ by studying $\{g^{\circ n}(z)\}_{n=1}^{\infty}$ because $$f^{\circ n}(z) = \varphi^{-1}_2(g^{\circ n}(\varphi_2(z)))$$
where $$\varphi_2(w) = w + \frac 1 w$$ for some appropriate domain and range
From where did $\varphi_2(w)$ come? It doesn't seem to be in the form $\varphi_1(z) = az + \frac b 2$, though I'm thinking there's some substitution to be done (hence the use of $w$ and not $z$)

I think you just have a mix up in the notations for $\varphi_2(w) = w + \frac{1}{w}$. Do not use the $w$ variable, stick to just $z$. So assume the conformal change of variables is
$$\varphi_2(z) = z + \frac{1}{z}$$
Assume $g(z) = z^2$ and $f(z) = z^2 - 2$. Then you can show that $$\varphi_2  \circ g = f \circ \varphi_2$$ This is equivalent to $$f = \varphi_2 \circ g \circ \varphi^{-1}_2$$
Let's check $\varphi_2  \circ g = f \circ \varphi_2$. It is enough to do that and everything else follows from it. Thus we have to verify the identity
$$ \varphi_2 \big( g(z)\big) = \varphi_2( z^2 ) = f\big(\varphi_2(z)\big) = \big(\varphi_2(z)\big)^2 - 2$$ Indeed 
$$ \varphi_2 \big( g(z)\big) = \varphi_2( z^2 )  = z^2 + \frac{1}{z^2} $$
$$f\big(\varphi_2(z)\big)  = \big(\varphi_2(z)\big)^2 - 2 = \left(  z + \frac{1}{z}\right)^2 - 2  = z^2 + 2 \, z \, \frac{1}{z} + \frac{1}{z^2} - 2 =$$ $$= z^2 + 2  + \frac{1}{z^2} - 2   = z^2 + \frac{1}{z^2}$$
