Here's an example that might make it clearer. Let's say that $X$ is the set of points in some square, and let's say that $G$ is the group $Z_4 = \{0, 1, 2, 3\}$. The group action of $G$ acting on $X$ will be that if $x$ is some point of the square, then $\phi(n, x)$ is the point on the square that you get after rotating $x$ counterclockwise around the center of the square by $n\cdot90°$.
So for example the group element $1$ acts on points of the square by rotating them a quarter-turn clockwise around the center of the square, and the group element $3$ acts on points by rotating them a quarter-turn counterclockwise.
The identity element for $G$ is 0, and indeed, we have $\phi(0, x) = x$ for each $x$ in the square, since the result of rotating $x$ around the center by 0° is $x$ again.
Similarly, we want $\phi(a, \phi(b, x)) = \phi(a+b, x)$ for each $x$ in the square and each $a,b$ in $Z_4$. This just says that if we rotate first by $b\cdot90°$ and then by $a\cdot90°$ that's the same as doing a single rotation by $(a+b)\cdot90°$, which is correct. So this is indeed an action of $G=Z_4$ on the square $X$.
This is just an example—there will be many other examples that are quite different—but it's a very typical example.
One thing you should notice is that the choice of a square and $Z_4$ match up closely here.
For example, suppose we took $Z_3$ instead of $Z_4$, but we say that $n$ still represents a rotation of $n\cdot90°$. In $Z_3$ we have $1+2 = 0$, so we should have that $\phi(1, \phi(2, x)) = \phi(1+2, x) = \phi(0, x)$ for any $x$ in the square. But this is wrong. Say $x$ is the upper right corner of the square. Then $\phi(2, x)$ is the lower left corner, and $\phi(1, \phi(2,x))$ is the lower right corner. But $\phi(0, x)$ is the upper right corner, not the lower right corner, so this is not an example of a group action.