In tensor notation, we know the following is true for general vectors:
$$ \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = A_i\epsilon_{ijk}B_jC_k = -B_j\epsilon_{jik}A_iC_k $$
However, if we try and introduce $\nabla$ and start thinking about vector calculus, we cannot be so free with manipulation. For instance, the same movement of terms as the above is no longer valid:
$$ \nabla\cdot (\mathbf{B} \times \mathbf{C}) = \partial_i\epsilon_{ijk}B_jC_k \neq -B_j\epsilon_{jik}\partial_iC_k = -\mathbf{B} \cdot (\nabla \times \mathbf{C}) $$
I am not able to formally explain why such a breakdown in the use of tensor notation when extended to differential vector operators occurs. What is the nature of $\nabla$ when it appears in tensor notation. Is there a general rule/ structure that must be followed?
In the second case, if you're treating $\mathbf{B}, \mathbf{C}$ as functions $\Bbb{R}^3\to \Bbb{R}^3$, then you can't just make the swap because differential operators (like $\partial_i$) follow a product rule, so that \begin{align} \nabla\cdot (\mathbf{B} \times \mathbf{C}) &= \partial_i\left(\epsilon_{ijk}B_jC_k\right) \\ &=\epsilon_{ijk} \partial_i(B_j C_k) \\ &= \epsilon_{ijk}\left[ (\partial_iB_j)\cdot C_k + B_j(\partial_i C_k)\right] \tag{$*$} \\ &= [\epsilon_{ijk} \partial_i B_j]\cdot C_k - B_j [\epsilon_{ikj}\partial_i C_k] \\ &= [\nabla \times \mathbf{B}]_k \cdot C_k - B_j [\nabla \times \mathbf{C}]_j \\ &= (\nabla \times \mathbf{B})\cdot \mathbf{C} - \mathbf{B} \cdot (\nabla \times \mathbf{C}) \end{align} Notice that step $(*)$ is where we used the product rule, and once again this is due to the fact that $\partial_i$ is a differential operator.
This issue doesn't arise in the case of your first formula \begin{align} \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = A_i\epsilon_{ijk}B_jC_k = -B_j\epsilon_{jik}A_iC_k \end{align} because at each stage we're dealing with the normal arithmetic of real numbers, which means all multiplication, addition etc are all associative and commutative (i.e you can bracket and rearrange terms as you wish).
Now, I do not know how you've been introduced to the $\nabla$ operator, but it can't hurt to emphasize this once again: $\nabla, (\nabla \cdot), (\nabla \times), \nabla^2$ are all differential operators which act on certain functions, respectively called gradient, divergence, curl, Laplacian (and sometimes the notation $\text{grad}, \text{div}, \text{curl}, \text{Lap} = \text{div}\,\text{grad}$ is also used). They are NOT vectors. The notation $\nabla$ has been chosen only to shorten the appearance of certain formulas and to evoke some kind of familiarity with the usual vector algebra. However, do not take this notation too seriously, because you can easily run into all sorts of troubles if you do.