Intuitive meaning of transitive action In the context of abstract algebra, following from Dummit's textbook. 
The action of $G$ on $A$ is called transitive if there is only one orbit. $i.e$, 
given any two elements $a,b \in A$, there is some $g \in G$ such that $a=g.b$. 
I want to know why we called this as transitive. 
From some basic set theory knowledge, 
In equivalence relation, transitivity, 
$a \sim b$, $b\sim c$ then $a\sim c$. 
Does the same thing happens in the group action? 
 A: Mathematicians like to copy names from different topics although their relation is ever so subtle. You should be aware of that because it can get confusing, especially when there is a generalization of a concept. 
A: To answer the title of your question, you can think of a transitive group action on a set $A$ as an action for which you can go from any point to another by the action of the group.
In general, this doesn't happen.  If $G$ is a group acting on a set $A$, and $a \in A$, let $Ga$ denote all the points of $a$ you can get to by applying elements of $G$.  Literally, $Ga = \{ g.a : g \in G \}$.  Then in general, you will have elements $a, a', a''$ etc. (possibly uncountably many) such that $A$ is the disjoint union of the sets $Ga, Ga', Ga''$ etc.  These sets are called the orbits of $A$ under the action of $G$.  If there is just one set in this disjoint union, that's transitive.
If $a \in A$, then you can always restrict the action of $G$ to the set $Ga$.  Then $G$ does act transitively on $Ga$.
Example:

1 . $G = \textrm{GL}_2(\mathbb R)$, the group of $2$ by $2$ real matrices with nonzero determinant.  $G$ acts on $\mathbb{R}^2 - \{(0,0)\}$ by $\begin{pmatrix} a & b \\ c & d \end{pmatrix} . (x,y) = (ax + by, cx + dy)$.  Since you can get from any given nonzero vector to another by applying some invertible matrix, the action here is transitive.
2 .  $G = \textrm{SO}_2(\mathbb R)$, the group of $2$ by $2$ real matrices $g$ such that the determinant of $g$ is $1$ and such that $gg^t = I$, where $g^t$ is the transpose of $g$ and $I$ is the identity matrix.  $G$ acts on $\mathbb{R}^2 - \{(0,0) \}$ by the same formula.  You can show every $g \in G$ takes the form $$g = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta &  \cos \theta \end{pmatrix}$$
for some $\theta \in \mathbb{R}$, and if $v \in \mathbb{R}^2$ is a nonzero, then $g.v$ is the vector obtained by rotating $v$ counterclockwise by $\theta$ radians.  In this case, you definitely cannot get from one given element to another by applying something in $G$, since you can only use $G$ to rotate points around.  So the action is not transitive.  It follows that the orbits of $\mathbb{R}^2 - \{(0,0\}$ under the action of $G$ are the circles with center $(0,0)$, and $\mathbb{R}^2 - \{(0,0)\}$ is the disjoint union of the orbits
$$\mathbb{R}^2 - \{(0,0)\} = \bigcup\limits_{r \in (0,\infty)}  \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = r^2 \}$$

