$y=a^x$
$\log_a(y)=\log_a(a^x)=x$

We knew that the divergence of curl is alwasy zero (provided the function is not pathological to disallow interchange of partial derivatives). i.e.
$$\nabla \cdot (\nabla \times \vec{V})$$

notions of derivative require certain "local" smoothness. if you are looking at a function that explodes to infinity at some point, well then curl grad etc cannot be defined at this point.
but if the point is isolated and the function smooth enough on the surrounding areas, then near any point you can find an open set which does not contain your singularity. so the function will be locally well behaved and div curl = 0 will hold everywhere except where these thigns are not defined

What surface is $B_1$? Cause any attempt I tried to pictorially and mentally piece a vector field F that give the above radial curl seemed to have all those microscopic circulations cancelling each other out on any closed surface, or is this a bad way to understand what happened?

Or in short, how can one have a curl of a vector field that is radial without all microscopic circulations represent by curlF anywhere to cancel out?

(I am really bad at analysis intuitions, sorry for that...)

The situation is easier in $2$ dimensions, where you have something like the gradient:
$$f(x,y)\mapsto \begin{pmatrix}\partial_x f\\ \partial_y f\end{pmatrix}$$
and something like a curl that goes to functions:

You can check quickly that the composition of the two maps is always zero. There is also an analogue of the Gauß theorem:
$$\int_A (d \vec{v})\ dxdy = \int_{\partial A} \vec{v}\cdot d\gamma$$

However if we calculate this integral, we will see the following:
$$\int_0^{2\pi} d\phi \begin{pmatrix} -\sin(\phi)/1\\ \cos(\phi)/1 \end{pmatrix}\cdot \begin{pmatrix}-\sin(\phi)\\ \cos(\phi)\end{pmatrix} = \int_0^{2\pi}d\phi (\sin^2+\cos^2) = 2\pi$$

How does the hole at $\{0\}$ contribute to the curl?

, I only recall back in class they have show this example and said that if the domain of integration is not simply connected then (forgot) will not hold and a line integral in a radial vector field will not be conservative

This is a sort of "moral" or "global" violation of $\mathrm{curl}\ \mathrm{grad}=0$, since we can also do the following:
$$\int_{S^1} df d\gamma = \int_{S^1} df d\gamma - \int_{S^1_\epsilon} df d\gamma + \int_{S^1_\epsilon} df d\gamma = \int_{B_1 - B_\epsilon} ddf + \int_{S^1_\epsilon} df d\gamma$$