Big-Oh and Growth Rate
You understood it correctly, but have interpreted it differently than the book wanted it to be interpreted. You are right that $9n$ grows faster than $n$ so the main problem in your book may be trying to express a mathematical definition of $O$ with too few words.

The definition
The function $f(n)$ is in $O(g(n))$ if there exist $n_0 \in \mathbb{N}$ and a positive real number $c$ such that $f(n) \leq c \cdot g(n)$ for all $n \geq n_0$.  

Now it is clearer how $9n$ is in $O(n)$. Because there exist, say $c=10$ and $n_0=0$ such that $9n \leq c \cdot n$ for all $n \geq n_0$.

Why compare so loosely?
The problem is, if we would compare growth rates by exact function values, then two very similar functions may not end up in the same class. For example, with your "strict-O notation"; for any function $d &gt; 0$, $f+d$ would not be in the same class as $f$, and that is not very useful. So, we have to draw the line somewhere (pun intended). Also, we use asymptotic analysis for describing algorithm complexity dependant on the input size $n$, and it is often not possible to calculate computation steps that precise.

Other notations
But, there are other asymptotic notations except big-O so you might want to look at those. You will find that $\Theta$ is, for example, more restrictive than $O$.