Consider $f(x)=2x$. It is a straight line with constant slope $2$, and that's what $f'(x)=2$ tells us.
Now consider $f(x)=x^2$. $f'(x)=2x$ tells us that the slope of $f$ is changing at each point, and that change is behaving like the function (straight line) $2x$. Then $f''(x)=2$ is telling us that the variation of the change of $f'$ is constant, which means that the slope of $f$ is changing, but it changes at a constant rate.
And yes, $f''$ does tell you how the slope of $f$ is varying, since the slope is given by $f'$ and the first derivative of $f'$ is $f''$. In an attempt to provide some intuition I'll use the following analogy:
Imagine $f$ as something that given a point $x$ gives you a position for $x$ on the graph. Then $f'$ is telling you how fast or slow a position is changing into the following positions (if $f'$ is positive it is changing upwards, and if $f'$ is negative it is changing downwards). So $f'$ could be though as "the speed" of the position moving. Then $f''$ is telling you how that change varies. If $f''$ is positive and big then $f'$ ("the speed") is going to get bigger rapidly, which means the following positions are going to change faster ($f''$ works as an "acceleration"). If $f''$ is negative then your $f'$ is going to get slower, so the position will change more and more slowly.