The diagram is simply a way to remember two variables at the same time, one called the total spin $\hbar\sqrt{3}/2$, the other called the z-component which has two possibilities $\pm\hbar/2$. If you draw a picture of a vector of length $\hbar\sqrt{3}/2$ that is at an angle so that the z-component of the vector is $\pm\hbar/2$ (so it lies on two cones, one with $+\hbar/2$ and one with $-\hbar/2$) the picture can help you remember those two numbers (and match them to the right name).
The names are very misleading however, so a picture to help you remember does not help much, and probably hurts more than it could ever help. That's because if you "measure" the component in any direction whatsoever you always get $\pm\hbar/2$, so clearly you aren't measuring a pre-existing component of some preexisting vector. If there were a preexisting vector what would happen if you measured the component orthogonal to that vector? You can't get zero since you always get $\pm\hbar/2$. What you are really doing is polarizing the spin so that later it will give the same result on future measurements as what you just got. So the spin direction should be telling you how it is polarized, specifically a direction that gives $+\hbar/2$ (and where if you measure in the 100% opposite direction for your z-axis you get $-\hbar/2$) and where if you measure in any other directions besides those two you get a statistical spread of answers, sometimes $+\hbar/2$ and sometimes $-\hbar/2$. The real physics is learning how to get that statistical spread, and that's what a quantum theory course would cover.
And since you always get $\pm\hbar/2$ for any component, that is exactly why the total spin is $\hbar\sqrt{3}/2$: $S=\sqrt{S^2}=\sqrt{S_x^2+S_y^2+S_z^2}=\sqrt{(\pm\hbar/2)^2+(\pm\hbar/2)^2+(\pm\hbar/2)^2}=\sqrt{3\hbar^2/4}=\hbar\sqrt{3}/2.$ There is nothing but harm coming from thinking there is a vector that points in a certain direction and you measure an already existing component of that vector.