In Cartesian coordinates, vector operations are as simple as if we were treating the del operator as like a vector.
$$\nabla = \Big(\frac{\partial}{\partial x}\hat{\textbf{i}} + \frac{\partial}{\partial y}\hat{\textbf{j}} +\frac{\partial}{\partial z}\hat{\textbf{k}}\Big)$$
For a vector field: $U = u\hat{\textbf{i}} + v\hat{\textbf{j}} + w\hat{\textbf{k}}$
$$\nabla \cdot \Big(u\hat{\textbf{i}}+v\hat{\textbf{j}}+w\hat{\textbf{k}}\Big) = \Big(\frac{\partial u}{\partial x}|\hat{\textbf{i}}||\hat{\textbf{i}}|\cos(0) + \frac{\partial y}{\partial y}|\hat{\textbf{j}}||\hat{\textbf{j}}|\cos(0) +\frac{\partial w}{\partial z}|\hat{\textbf{i}}||\hat{\textbf{i}}|\cos(0)\Big)$$
$$\nabla \times \Big(u\hat{\textbf{i}}+v\hat{\textbf{j}}+w\hat{\textbf{k}}\Big) = \matrix{|\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}| \\ |u & v & w |} = \frac{\partial v}{\partial x}(\hat{\textbf{i}}\times\hat{\textbf{j}}) + \frac{\partial u}{\partial y}(\hat{\textbf{j}}\times\hat{\textbf{i}}) + \frac{\partial w}{\partial y}(\hat{\textbf{j}}\times\hat{\textbf{k}})+\frac{\partial v}{\partial z}(\hat{\textbf{k}}\times\hat{\textbf{j}}) \\ +\frac{\partial u}{\partial z}(\hat{\textbf{k}}\times\hat{\textbf{i}})+\frac{\partial w}{\partial x}(\hat{\textbf{i}}\times\hat{\textbf{k}})$$
and so on. It behaves very much like a vector in Cartesian coordinates. The same is true when we move to cylindrical coordinates:
$$\nabla = \Big(\frac{\partial}{\partial r}\hat{\textbf{r}} + \frac{1}{r}\frac{\partial}{\partial \phi}\hat{\textbf{$\phi$}} +\frac{\partial}{\partial z}\hat{\textbf{k}}\Big)$$
For a different vector field: $V = u\hat{\textbf{r}} + v\hat{\textbf{$\phi$}} + w\hat{\textbf{k}}$, that is the components u, v, and w point purely in the radial, azimuthal, and vertical directions
$$\nabla \cdot \Big(u\hat{\textbf{r}}+v\hat{\textbf{$\phi$}}+w\hat{\textbf{k}}\Big) \ne \Big(\frac{\partial u}{\partial r}\hat{\textbf{r}} + \frac{1}{r}\frac{\partial v}{\partial \phi}\hat{\textbf{$\phi$}} +\frac{\partial w}{\partial z}\hat{\textbf{k}}\Big)$$
What this is instead supposed to equal is: $\Big(\frac{1}{r}\frac{\partial u}{\partial r}\hat{\textbf{r}} + \frac{1}{r}\frac{\partial v}{\partial \phi}\hat{\textbf{$\phi$}} +\frac{\partial w}{\partial z}\hat{\textbf{k}}\Big)$. I don't understand how exactly the Del operator behaves here that leads to the new $\frac{1}{r}$ term in the result. I've seen several answers on this topic bring up tensor analysis and relevant notation, but I don't know much about this area of mathematics, I mostly only know about basic Calculus I-III. Is there an explanation for how it works that is contained within introductory Calculus?
