The main point to note is that when we do the general derivation for work energy theorem, we do a definite integral. When we do a definite integral of a function over a variable, the result is an expression independent of that variable. Consider for eg:
$$y(a,x)= ax$$
Where $a$ is a constant, if we integrate from $x=0$ to $x=1$, we have:
$$ G(a)= \int_{x=0}^1 y(a,x) dx = a \frac{x^2}{2}|_{x=0}^{x=1} = \frac{a}{2}$$
Similarly in the work energy theorem, we can simplfy the expression to result in an integral with time.
$$ \int F \cdot dx = \int \left( ma \cdot v \right) dt $$
So, clearly when we do the last integral over some bounds, we arrive at an expression which would have no "time" in it. Note that the simplification I did of:
$$ dx =v dt$$
Is not reliant on the thing I am integrating being $ma$, it could have been any expression for force.
Example
Consider particle falling down,
$$ ma = -mg \hat{z}$$
we have:
$$ \int_{t=t_0}^{t=t_f} ma \cdot dx = \int_{t=t_0}^{t=t_f} -mg \hat{z} \cdot dx$$
$$ m \frac{\left(v(t_f) \right)^2}{2}- m \frac{\left(v(t_o) \right)^2}{2} = -mgh(t_f) - mgh(t_o)$$
You can see in the expression above that time is plugged in into the functions already. This is what the paragraph tells us.
Additionally, what we can do to make this more useful is to rearrange all the terms which are evaluated at $t_f$ to one side, we have:
$$ E= m \frac{\left(v(t_f) \right)^2}{2}+ mgh(t_f) = m \frac{\left(v(t_o) \right)^2}{2}+ mgh(t_o)$$
So, our conclusion in this case would be that the sum of potential and kinetic energy evaluated at a time $t$ is same for all time $t$.