Does the rate of turn increase or decrease by turning at a higher speed?
The first formula is a function of bank angle $\phi$ and the velocity $v$. By varying only $v$, you are keeping $\phi$ constant. As a consequence, to keep the turn "steady" and coordinated (that are the assumptions needed to write that formula) you need to increase the radius of turn $r$ and decrease the turn rate.

The second formula is a function of $r$ and $v$. Once again you change only $v$, $r$ will be constant here (while before was $\phi$). This leads to a turn with a higher $\phi$ (again to mantain valid the assumptions behind the formula) and higher rate of turn.

So, to conclude, assuming that you whish to perform a coordinate turn, whether the turn rate increases or decreases is a function of what you whish to keep constant when you change speed:


if you have a constraint on the bank angle, an increase of the velocity will decrease the turn rate and increase the turn radius
if you have a constraint on the radius, an increase of the velocity will increase both bank angle and turn rate


To complete the set of equations and see this in practice, you would need to flip each equation and apply it after the other. That means:


Case 1:


$$\dot\psi=\frac{g \cdot tan\phi}{v}   \;\;\;,\;\;\; r = \frac{v}{\dot\psi}  \;\;\;,\;\;\; \phi = const.$$


Case 2:


$$\dot\psi=\frac{v}{r} \;\;\; , \;\;\; \phi = atan\left( \frac{v \cdot \dot\psi}{g} \right) \;\;\; , \;\;\; r= const. $$

Note in case 2 how $\phi$ is computed through an $atan()$ function. This means that increasing $v$ will make $\phi$ approach $90^\circ$. The validity of the model used to derive this equations is no more guaranteed at that point.