The second part, $ \newcommand{PYL}{\operatorname{P\Gamma L}} \newcommand{PEL}{\operatorname{P\Sigma L}} \newcommand{PGL}{\operatorname{PGL}} \newcommand{PSL}{\operatorname{PSL}} \newcommand{Aut}{\operatorname{Aut}} \newcommand{Alt}{\operatorname{Alt}} \newcommand{Sym}{\operatorname{Sym}} \left\{ f \in \PYL(2,q^2) ~\middle|~~f:z\mapsto\frac{az^q+b}{cz^q+d},0\neq ad-bc\neq k^2\right\}$ is not a group, just a coset. Let's pretend $q$ is prime so that I don't have to make up any non-standard notation. Divide $\PYL(2,q^2)$ into cosets over $\PSL(2,q^2)$. The cosets have representatives $1$, the Frobenius automorphism $\sigma:z\mapsto z^q$, the diagonal element $\tau:z \mapsto \zeta z$ where $\zeta$ is a primitive $q+1$st root of unity, and of course $\sigma\tau$ the combination of the last two.
The group $\PYL(2,q^2)/\PSL(2,q^2)$ is elementary abelian of order 4, and so it has three non-identity proper subgroups, each generated by a single element: $\sigma$, $\tau$, or $\sigma\tau$.
Using the lattice isomorphism theorem (subgroups of a quotient correspond to subgroups of the original containing the kernel), we get the following subgroups:
$\begin{array}{rcl} \PEL(2,q^2) & = & \PSL(2,q^2) \cup \sigma\PSL(2,q^2) \\ \PGL(2,q^2) & = & \PSL(2,q^2) \cup \tau\PSL(2,q^2) \\ M(q^2) & = & \PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2), \text{ as well as} \\ \PSL(2,q^2) & = & \PSL(2,q^2) \text{ and } \\ \PYL(2,q^2) & = & \PSL(2,q^2) \cup \sigma\PSL(2,q^2) \cup \tau\PSL(2,q^2) \cup \sigma\tau\PSL(2,q^2) \end{array}$
When $q=3$ we get a particularly important version of this that you'll want to know about at some point:
$\begin{array}{rcl} \PSL(2,9) & \cong & \Alt(6) \text{ the alternating group of degree 6 } \\ \PEL(2,9) & \cong & \Sym(6) \text{ the symmetric group of degree 6 } \\ \PGL(2,9) & = & \PGL(2,9) \\ M(9) & \cong & M_{10} \text{ the Mathieu group of degree 10 } \\ \PYL(2,9) & \cong & \Aut( \Alt(6) ) \text{ the automorphism group of the alternating and symmetric groups } \end{array}$
I believe the "M" is not an abbreviation for "Mathieu". Huppert–Blackburn (Vol 3, XI.1.3 p. 163) does not name it.