Revised answer (see previous version to make sense of comments):
The Petersen graph is a small wonderful and rather exceptional creature. It is not a paradigm.
VaryDoes this graph fit your construction:
For with $G = C_m$$G = C_7$, $H = K_2$ and?
- $\pi_1(i) = \text{id}$ for $i=1,\dots,m$
- $\pi_2(1) = \text{id}$
- $\pi_2(2) = (12)(3\cdots m)$ link text I don't think that the result is vertex transitive so your construction need not (and probably seldom) gives a vertex transitive result.
It is not vertex transitive since each red node is in three $5$-cycles but each green node is only in two.
There are many vertex transitive graphs with a prime number of vertices, none of them is a product. So there are graphs which do not arise in this manner.
WhatYour construction takes two graphs can you get? Can you get$G_i$ with $K_{2m}$ minus$v_i$ vertices each of degree $d_i$ and some permutations and creates a matchingnew one with $v_1v_2$ vertices each of degree $d_1+d_2.$ This can work out to be a rather low degree (or high if a complement is taken.) Consider the Johnson graph $J(6,2)$ whose $\binom{6}{2}=15$ nodes are the pairs from an $6$ set with two adjacent when they share an element. It has degree $2(6-2)=8.$ If it is to be a product then we need $v_1=3$ and $v_2=5$ which allows for degree at most $m=3$, or$2+4=6.$ This also does not allow the product to be the complement a regular graph of degree $7.$ A similar obstacle would arise for many vertex transitive graphs just based on the number of vertices and the degree. Certainly for $J(n,2)$ when $n \ge 6$ makes $\binom{n}2$ odd and probably even anywhen $m \ge 3$?$\binom{n}2$ it is even.