Yes, they form a basis.

Great! Thanks for the quick answer!

Cojugating by a commuting element does not give a new automorphim. It is unclear to me that conjugating shall give infinitely many different automorphisms.

What is the Picard number of this K3 surface?

@Evgeny Shinder, for your first idea: Yes. In fact, if the composite $\sigma_1 \circ \sigma_2$ is of finite order, then the property a) is satisfied.

@Evgeny Shinder, the subgroup $H$ does not need to be finite and so you don't necessarily have a surface $X/H$. In fact, one can find many examples of K3 surfaces with $\sigma_1, \sigma_2$ that satisfy the property b) only. I just added a comment in the main post about this.