On
$\dfrac{1}{r}\begin{pmatrix} \vec{e_r} & r \vec{e_ \theta} & \vec{e_z} \\ \dfrac{\partial}{\partial r} & \dfrac{\partial}{\partial \theta} & \dfrac{\partial}{\partial z} \\ V_r & r V_\theta & V_z \end{pmatrix} = \bigg(\dfrac{1}{r} \dfrac{\partial V_z}{\partial \theta} - \dfrac{\partial V_\theta}{\partial z} \bigg) \vec{e_r} + \bigg(\dfrac{\partial V_r}{\partial z} - \dfrac{\partial V_z}{\partial r} \bigg) \vec{e_\theta} + \dfrac{1}{r} \bigg( \dfrac{\partial (r V_\theta)}{\partial r} - \dfrac{\partial V_r}{\partial \theta} \bigg) \vec{e_z}$
Why in $\vec{e_z}$, ${\dfrac{\partial (r V_\theta)}{\partial r}}$ isn't simplified by $\frac{1}{r}$ as it is in $\vec{e_r}$ cordinate? (that was $\dfrac{\partial r V_\theta}{\partial z}$ before)
Your question is a little unclear but I think the answer is as follows: you've misunderstood how $\frac{1}{r}$ cancels things out. By the product rule, you have
$$ \frac{\partial}{\partial z} (r V_\theta) = r \frac{\partial V_\theta}{\partial z} + V_\theta \frac{\partial r}{\partial z}$$ $$ \frac{\partial}{\partial r} (r V_\theta) = r \frac{\partial V_\theta}{\partial r} + V_\theta \frac{\partial r}{\partial r}$$
While partial derivative notation is generally ambiguous, it's clear here from the use of the coordinate chart $(r, z, \theta)$ that $\frac{\partial}{\partial z}$ meant to be the derivative as $r$ and $\theta$ are held constant, and similarly that $\frac{\partial}{\partial r}$ is meant to be the derivative as $z$ and $\theta$ are held constant. So,
$$ \frac{\partial}{\partial z} (r V_\theta) = r \frac{\partial V_\theta}{\partial z} + V_\theta \cdot 0 = r \frac{\partial V_\theta}{\partial z}$$ $$ \frac{\partial}{\partial r} (r V_\theta) = r \frac{\partial V_\theta}{\partial r} + V_\theta \cdot 1 = r \frac{\partial V_\theta}{\partial r} + V_\theta$$
So this is why the $r$ "factors out" (so that it may be cancelled by $\frac{1}{r}$) in the partial derivative with respect to $z$. The derivative with respect to $r$ is not so simple.