Position is $(x(t), y(t))$ and velocity is $v(t) = (\dot{x}(t), \dot{y}(t))$, acceleration is $a=\dot{v} = (\ddot{x}, \ddot{y})$, and speed is the scalar function $s(t) = |v(t)| = \sqrt{\dot{x}^2 + \dot{y}^2}$.
I am using the dot notation, $\dot{u} = u' = \frac{du}{dt}$, because it is easier to read with exponents.
So let's use the chain rule:
$$ \dot{s} = \frac{1}{2\sqrt{\dot{x}^2 +\dot{y}^2}} \bigg( 2\dot{x}\ddot{x} + 2\dot{y}\ddot{y} \bigg) = \frac{\dot{x}\ddot{x} + \dot{y}\ddot{y}}{\sqrt{\dot{x}^2 +\dot{y}^2}}$$
If you are familiar with dot products, this is:
$$ \dot{s} = \frac{ v\cdot a }{s} = \left(\frac{v}{|v|}\right)\cdot a $$
Notice: $v/s = v/|v|$ is the unit vector in the direction of motion. The dot product of a vector with a unit vector gives the component of the vector in the direction of the unit vector. So this means:
The derivative of speed is the component of acceleration in the direction of motion.
So let's look at your graph. The vertical axis is indeed the derivative of speed.
In projectile motion, a ball is shot upward. Its speed is initially declining as gravity acts on it, so the derivative of speed is negative. As the ball reaches its apex, its motion becomes horizontal, perpendicular to gravity, so its speed doesn't change so much; at the apex, the ball's speed is constant, and that's the instant that $\dot{s}$ passes through $0$.
Once the ball is falling downward, its speed increases: positive $\dot{s}$, and as its motion becomes more and more downward, its speed derivative asymptotically tends toward the magnitude of gravity's acceleration.