The problem with this question, compared to this one of yours, is that the vector field on the right hand side of the ODE is not a non-decreasing function of $g$. If you try to make the example of @fedja rigorous, you can see that you can manage to build even smooth counterexamples. To make everything work, you need an ODE of the form $g’(t)=v(g(t))$, where $v$ is non-decreasing.

Another example, this time with a positive vector field:
$$ g’(t)=g(t)(2-g(t)), $$
with initial data $f(0)=g(0)=1$. Then
$$ g(t)=\frac{2e^{2t}}{e^{2t}+1}. $$
The corresponding integral inequality
$$ f(t)\leq 1+\int_0^t f(s)(2-f(s))ds $$
admits as a solution, for instance,
$$ f(t)=\frac{1+2e^{-(x-10)^2}}{1+2e^{-(10)^2}}. $$
Clearly, $f(10)=3>2>g(10)$. The problem is that the vector field $v(x)=x(2-x)$ fails to be non-decreasing in its argument.

Heuristic explanation. The second term of your integral equation is like a ‘bag’ that saves you a quantity of energy $v(f(s))$ per second. The inequality you want to prove is disproved as soon as you make $f$ gain more total energy than $g$.

If $v$ is a decreasing function, then as soon as $f(t)<g(t)$, you have $v(f(t))>v(g(t))$. That is, $f$ collects more energy than $g$ per second. So to make a large $f$, it is convenient to keep $f$ small for a suitable amount of time (so that it satisfies the inequality) and at the same time make it stay in a favorable range where you can collect more energy (in my previous example, $v$ is maximized at $x=1$), and use that energy later to grow past $g$ as soon as you have collected enough energy (while $g$ is forced to grow by the ODE and as soon as it gets large it will start collecting energy with a smaller rate with respect to $f$).

how to show...", not "is it true that...". Besides, this post hardly passes the criteria for "being appropriate for MO", so a full answer would not be a right thing to post. I just did what I usually do in such situations: left a helpful comment and let other people decide if they want to close the thread :-) $\endgroup$1more comment