I dont understand the G syntax in a LTL (linear temporal logic) formula I know it states: "G for always (globally)"
But what does this mean? Is this the "same" as A for CTL syntax?
What is the difference between 
M |= AG EF p (this i read as globally for all paths there exists a path where evenutally p is true)
and
M |= A EF p
It seems that G in LTL is very similar to A in CTL
 A: $\mathsf{G}$ is a temporal operator (or modality). $\pi \models \mathsf{G} p$ means that $p$ holds at all states of path $\pi$.
$\mathsf{A}$ is a path quantifier.  In CTL and CTL$^*$, $\mathsf A$ quantifies over all the paths originating from a state.  In LTL it is as if there were an implicit $\mathsf A$ in front of the whole formula.  In fact, to translate from LTL to CTL$^*$, one simply adds an $\mathsf{A}$ in front of the LTL formula.
The first example you gave, $M \models \mathsf{AG\,EF} \,p$, concerns a CTL formula that says that from all states of $M$ reachable from the initial states of $M$ there originates a path along which $p$ eventually holds.  (This property is often called resetability, because $p$ may be chosen to distinguish the reset states of the model.)
The second example you gave, $M \models \mathsf{A\,EF} \,p$, concerns a CTL$^*$ formula equivalent to $\mathsf{EF} \,p$.  $M$ satisfies $\mathsf{EF} \,p$ if, from all initial states of $M$, a state where $p$ holds is reachable. 
Neither example is expressible in LTL.  Both require branching time.
Perhaps, the CTL (and CTL$^*$) formula $\mathsf{AG} \,p$ illustrates the difference between $\mathsf{A}$ and $\mathsf{G}$ best.  In English, $M,s \models\mathsf{AG} \,p$  says "along all states of all paths of $M$ originating from state $s$, $p$ holds."  Both $\mathsf{A}$ and $\mathsf{G}$ are necessary in CTL to express that $p$ is invariant in $M$.
In LTL one simply skips the initial (implicit) $\mathsf A$, because the definition of $M \models \varphi$, when $M$ is a Kripke structure, incorporates the universal quantification over the paths originating from the initial states of $M$.

Consider a Kripke structure $M$ with states $\{0,1,2\}$, initial states $\{0\}$, and the following transition relation,
$$ \{(0,1),(0,2),(1,2),(2,2)\} \enspace. $$
Suppose that the atomic proposition $p$ holds at states $0$ and $2$.  Then we have $M \models \mathsf{F} p$, but $M \not\models \mathsf{G} p$.  If $\pi_1$ is the path $0,1,2,2,\ldots$ and $\pi_2$ is the path $0,2,2,\ldots$, then $\pi_2 \models \mathsf{G} p$, but $\pi_1 \not\models \mathsf{G} p$; hence $M \not\models \mathsf{G} p$.
