$x(t)$ simply means the value of the x co-ordinate at time $t$.
Draw a diagram of the x axis on which the object is moving. $x$ increases (becomes more +ve) to the right and decreases (becomes more -ve) to the left.
$dx/dt=v<0$ means that the x co-ordinate is decreasing as time passes. $x$ is becoming more -ve, the object is moving left. If it is to the right of O (ie $x>0$) then it will be moving towards O. However, if it is to the left of O ($x<0$) then the object will be moving away from O.
Likewise for $dx/dt=v > 0$. This means that $x$ is increasing, the object is moving right. If the object is left of O it will be moving towards O but if it is already to the right of O it is moving away from O.
$x^2$ represents the squared distance from O. It is always positive so it does not have the above complications with signs. If $d(x)^2/dt<0$ then the (squared) distance from O is always decreasing, whether the object is to the right or left of O. So it is moving towards O.
Likewise for $d(x)^2/dt>0$. The distance from O is increasing : the object is getting further from O, whichever side of O it is already.