Does the second derivative tell me how the variation of the slope of the tangent line to the graph varies? The first derivative tells me how the slope of the tangent line of each point to the graph varies, while the second derivative tells me how it varies as a true variation?
For example, having $f(x) = x^2$, $f'(x)= 2x$ (therefore the slope of each point of the line tangent to the point is equal to $2x$, for example at the point $x = 1$ is equal to $2$).
While the second derivative tells me how this variation varies, $f ''(x) = 2$. So $f''$ what are you telling me? I don't understand why it's a constant if the slope changes point to point.
 A: 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.
