The link you posted is about time-invariance of the dynamical system, meaning that the behavior of the system at some time $t$ is the same as at some other time $t′$.
For instance, the dynamical system $\dot{x}(t)=ax(t)$ where $a\in\mathbb{R}$ is time invariant. If you start the system at time $t_0$ from the state $x(t_0)=x_0$ and if you start the system at time $t_1\ne t_0$ from the state $x(t_1)=x_0$, then the two trajectories will be translated versions of each other.
The dynamical system $\dot{x}(t)=tx(t)$ is not time-invariant, but time-varying. In this case, the initial time will matter.
The composition formula
$${\displaystyle \Phi (t_{2},\Phi (t_{1},x))=\Phi (t_{2}+t_{1},x),}$$
works for both time-varying and time-invariant systems (with no input).
Finally, the example you gave is indeed a dynamical system whose state is given by $x(t)=x(t_0)+(t-t_0)^2/2$. In this case, the system has an input and this is not captured by the $\Phi$ formula you gave which is for systems with no input. In fact, the solution to that system is given by
$$x(t)=\Psi(t,s)x(s)+\int_s^t\Psi(t,\theta)\theta d\theta,$$
where $\Phi(t,s,x)=\Psi(t,s)x=x$, $t\ge s$, and where $\Psi(t,s)$ is the so-called state transition matrix (see e.g. https://en.wikipedia.org/wiki/State-transition_matrix). Finally, since the system is time-varying, the trajectory will not only depend on the initial condition but also on the initial time.
If you want to know more about dynamical systems, you may look at "Nonlinear Dynamics and Chaos" by Steven Strogatz. It is usually a good starting point.
Best Answer
Staying at the fixed point is surely one possible solution of the ODE. But to conclude that the system must stay at the fixed point, you need to know that the solution is unique. (The standard sufficient condition for this is that the right-hand side is a Lipschitz continuous function of $x$; see the Picard–Lindelöf theorem.)
Your example is a classical example of an ODE with non-unique solution. (The right-hand side $2 \sqrt{x}$ is not Lipschitz at $x=0$.)