We say that a function $f(n)$ is $O(n^2)$ if there exist constants $C,N>0$ such that $f(n) \leq Cn^2$ for all $n \geq N$. We usually denote "$f(n)$ is $O(n^2)$" by "$f(n) = O(n^2)$".
An algorithm has running time $O(n^2)$ if its worst-case running time is $O(n^2)$. That is, if we denote by $T(n)$ the maximum running time of the algorithm on inputs of length $n$, then the algorithm has running time $O(n^2)$ if $T(n) = O(n^2)$.
Equivalently (and this is the point of the CLRS formulation), an algorithm has running time $O(n^2)$ if there exists some function $f(n) = O(n^2)$ such that the running time of the algorithm on an input of length $n$ is always at most $f(n)$.
If we denote the running time of the algorithm on the input $x$ by $T(x)$, and the length of $x$ by $|x|$, then we can see the difference between these two equivalent definitions by writing them out formally.
The first definition is
$$
\max_{|x| = n} T(x) = O(n^2),
$$
where the left-hand side defines a function of $n$, which we denoted above by $T(n)$.
The second definition states that for some $f(n) = O(n^2)$,
$$
T(x) \leq f(|x|).
$$
If the second definition holds, then in particular
$$
\max_{|x|=n} T(x) \leq f(n),
$$
and so
$$
\max_{|x|=n} T(x) = O(n^2),
$$
which is the first definition.
If the first definition holds, then we can take $f(n) = \max_{|x|=n} T(x)$ to obtain a function satisfying the second definition:
$$
T(x) \leq \max_{|y|=|x|} T(y) = f(|x|),
$$
where the first definition guarantees that $f(n) = O(n^2)$.
