Conjecture
If we have two consecutive prime numbers $p_{a}$ and $p_{a+1}$, and two other consecutive primes $p_n$ and $p_{n+1}$, so that $p_{a} < p_{a+1} < p^2_{n+1}$, then $p_{a+1} - p_{a} < 2p_{n} $.
Are there any known counter examples and are there any known similar conjectures?
No known counterexamples. A slightly stronger conjecture, true as far as anyone has been able to check (up to $4 \cdot 10^{18}$) is that $$ \unicode{x2E2E} \unicode{x2E2E} \unicode{191} \unicode{191} \; p_{a+1} < p_a + 2 \; \sqrt {p_a} \; \unicode{63} ? $$ This is currently unprovable.
What people actually suspect is that, $$ \unicode{191} ¿ \; \mbox{if} \; \; p_a \geq 11, \; \; \mbox{then} \; \; p_{a+1} < p_a + \log^2 {p_a} \; ? $$ Really, really beyond proof.
See Prime pair points slope approaches 1
Note: as far as using the extensive tables of prime gaps, there is a detail involving the fact that $\log^2 x > 2 \sqrt x$ for an interval, roughly $19.6 < x < 187.8.$ So, it was necessary to make a separate confirmation of my version of the conjecture for $p_a < 188.$