I mentioned in a comment that you need more requirements on $f$ than just that is continuous. To give you a verbal explanation of the theorem I will assume it is non-decreasing. Then you can look at it as follows:
Since $f$ is non decreasing, $f(a)$ must be the minimum of $f$ over the interval, and $f(b)$ must be the maximum. Now it must be true that:
$$\int_a^b f(x)g(x)dx \geq f(a) \int_a^b g(x) dx$$
$$\int_a^b f(x)g(x)dx \leq f(b) \int_a^b g(x) dx$$
Now consider the function $F$ of $c$ given by
$$F(c) = f(a)\int_a^c g(x)dx + f(b)\int_c^b g(x) dx$$
This function must satisfy $F(b) \leq \int_a^b f(x)g(x)dx$ and also
$F(a) \geq \int_a^b f(x)g(x)dx$. Since it is continuous there must be a $c^*$ where equality holds. (By the intermediate value theorem).
So to put it in words. If you integrate a function $g$ from $a$ to $b$ and weight it by an increasing function $f$, then the weighted integral must be greater than the integral of $g$ times $f$'s min and less than the integral times $f$'s max. So there must be a point in between where $f$s min times some of $g$'s integral plus $f$'s max times the rest of $g$'s integral equals the total weighted integral.