When we first learn the dot product, it is described as a straightforward operation on two vectors (thought of as lists of real numbers), defined by $(a_1,a_2,\dots,a_n)\cdot(b_1,b_2,\dots,b_n)=a_1b_1+a_2b_2+\cdots+a_nb_n$.
Later, we come learn that this "mechanical" definition has an abstract counterpart: It is a specific example of an inner product, which is a map $V\times V\to\mathbb R$ for some real vector space $V$ that satisfies a few axioms (symmetry, bilinearity, and positive definiteness). There's also a natural way to extend the notion to complex vector spaces.
Now consider how the gradient operator $\nabla$ is often defined in a "naive" sense similar to the first version of the dot product above: We think of $\nabla$ relative to $\mathbb R^n$ as a "vector of operators" $(\partial_1,\partial_2,\dots,\partial_n)$. This is at least a helpful mnemonic for remembering the various definitions of vector calculus operations involving scalar fields $f:\mathbb R^n\mapsto\mathbb R$ and vector fields $A:\mathbb R^3\to\mathbb R^3$:
\begin{align} \nabla f&=(\partial_1f,\partial_2f,\partial_3f)\\\\ \nabla\cdot A&=\partial_1A_1+\partial_2A_2+\partial_2A_2, \end{align}
and similarly for $\nabla\times A$ and $\nabla^2f$.
But is there a more abstract way to make sense of $\nabla=(\partial_1,\partial_2,\dots,\partial_n)$? In particular, by analogy with the more abstract definition of inner products and its generalization to complex vector spaces, I'd like an understanding of $\nabla$ that generalizes to complex-valued functions. For example, in quantum physics one often encounters "complex scalar fields" $f:\mathbb R^3\mapsto\mathbb C$, whose gradients are "complex vector fields" $\nabla f:\mathbb R^3\mapsto\mathbb C^3$. Now is the "$\cdot$" in $\nabla\cdot(\nabla f)$ most naturally interpreted as a real or complex inner product? In practice (in quantum physics), the answer appears to be a real inner product. But I suspect there is a purely mathematical perspective on why this must be the case. [Meanwhile $|\nabla f|^2=(\nabla f)\cdot(\nabla f)$ makes sense, and in this case the "$\cdot$" coincides with the complex inner product on $\mathbb C^3$.]
I think the basic issue here is that the "vector notation" you're using cannot unambiguously describe the operations involved. In tensor analysis, it's usually abandoned in favor of more descriptive forms such as abstract index notation. For instance, I would gerally interpret $|\nabla f|^2$ to mean something quite different from $(\nabla f)\cdot(\nabla f)$, since single bars $|\ \ |^2$ generally refer to the Hermitian product on $\mathbb{C}$, but the dot $\cdot$ seldom does (this can vary, of course, and it's ultimately up to each author to define the notation they use).
Mathematically, the issue is that you are dealing with tensor fields over $\mathbb{R}^3$, and while $\cdot$ usually indicates some tensor contraction, there are often multiple ways to contract tensor fields.