Misunderstanding linearization of ODE : why $\dot y=\nabla f(\bar x(t))y+O(|y|^2)$ describe orbits near of $\bar x(t)$? Consider the ODE $$\dot x=f(x),\tag{*}$$
and let $\bar x(t)$ a solution. I'm reading the book "Introduction to applied Nonlinear Dynamical Systems and Chaos" by Wiggins. In the linearization part of page 7 they are really unclear. 

So, I want to understand the nature of solutions near $\bar x(t)$. What they do is : Let $x=\bar x(t)+y$. Substituting thins into $(*)$ and doing a taylor expension about $\bar x(t)$ gives $$\dot{\bar x}(t)+\dot y=\dot x=f(\bar x(t)+y)=f(\bar x(t))+\nabla f(x(t))y+O(|y|^2),$$
and since $\dot{\bar x}(t)=f(\bar x(t))$ we get $$\dot y=\nabla f(\bar x(t))y+O(|y|^2),$$
which describes the evolution of orbits near $\bar x(t)$.

Many question
1) In $x=\bar x(t)+y$ why $\dot y$ and $\dot x$ are not vanishing since they don't depend on $t$ ? There notation are very confusing, sometimes they use $x$ and sometime $x(t)$, so we never know what depend on $t$ and not depend on $t$. Also, why $\dot x=f(x)$ ? It's written nowhere that it's a solution. 
2) If $y$ describes orbits near $\dot x(t)$, shouldn't $y$ be also a solution of $\dot x=f(x)$ ? i.e. $\dot y=f(y)$ ? I don't understand why $y$ describes orbits near $x(t)$.
3) Why the solution of $\dot y=\nabla f(\bar x(t))y$ is a good approximation of $\bar x$ when $y$ very small ? (still is it a good approximation of the solution $\bar x$ or just of $\bar x(t)$ ?.
 A: Yes, it's a bit confusing.  I don't know if I can do better, but I'll try.
The equation is $\dot{x} = f(x)$.  It is understood that $x$ is a function of $t$.
Let $x = \overline{x}(t)$ be one solution of the equation.  That is, $\overline{x}(t)$ is a function such that $\dfrac{d}{dt} \overline{x}(t) = f(\overline{x}(t))$.
Let $x = \overline{x}(t) + y(t)$ be another solution of the same equation.  Thus
$$\dfrac{d}{dt} (\overline{x}(t) + y(t)) = f(\overline{x}(t) + y(t))$$
Thus $$\dfrac{d}{dt} y(t) = f(\overline{x}(t) + y(t)) - f(\overline{x}(t))$$
Since $f$ is nonlinear, the right side is not the same as $f(y(t))$.  However, considering $y(t)$ as small we have
$$f(\overline{x}(t)+y(t)) - f(\overline{x}(t)) = \nabla f(\overline{x}(t)) y(t) + O(|y(t)|^2)$$
In general $y(t)$ is not describing a solution of your equation, it's the difference between two nearby solutions.  However, one important special case is  that $\overline{x}(t)$ is constant (an equilibrium solution), so after a change of origin to make $\overline{x} = 0$, $y(t)$ does describe a solution.
