What is the rule for using $| \cdot |$ and $\| \cdot \|$ in Cauchy-Schwarz inequality
There are three separate issues:
Using $\|v\|$ or just $|v|$ to express the norm of a vector.
Using an absolute value on $\langle u, v \rangle$ or not.
Using $\langle u, v \rangle$ or $u \cdot{} \ v$.
The first is really a matter of style. What speaks in favor of $\| v\|$ is to highlight that it is the norm of a vector, what speaks in favor of $|v|$ is that that $|u\dot \ v| \le |u| \ |v|$ makes the formulation really slick, perhaps overly so.
Also the third is just a matter of style. Both mean the (or a) scalar product of $u$ and $v$.
For the second, for real vectors spaces in a strict sense it is weaker without the absolute value, yet the one with the absolute value can be derived easily from the one without (considering the inequality with $-u$ instead). Thus it is basically the same. For complex vectors though, it does not make sense to write an inequality without the absolute value.
In any case, the content is that the absolute value of the scalar product of two vectors $u$ and $v$ is bounded above by the product of the norms (induced by the scalar product) of the two vectors.
It's very simple. Whenever you have a real inner product space and define the norm of $x$ to be the square root of the scalar product of $x$ with itself, then you have the Cauchy-Schwarz inequality. Whatever is the notation for the inner product and for the norm.
Remark 1: The formula with the $\cos$ may be understood as the definition of the angle.
Remark 2: The C-S inequality $$\mathrm{|scal\_product|} \leq \mathrm{product\_of\_norms}$$ immediately implies $$\mathrm{scal\_product} \leq \mathrm{product\_of\_norms}$$
Give a real or complex vector space $V$ and a positive semidefinite symmetric sesquilinear form $\langle \cdot, \cdot \rangle$ on $V$ one can define the norm of of an element $v \in V$ as $\|v\| = \sqrt{\langle v, v \rangle}$. The Cauchy-Schwarz inequality then states that $$ |\langle v, w \rangle| \leq \|v\| \cdot \|w\| \quad \text{for all $v,w \in V$}. $$ Depending on the situation one can now use slightly different notations or state this in slightly different, but equivalent ways:
Instead of writing $\langle \cdot, \cdot \rangle$ or $\|\cdot\|$ one might prefer to use a simple $\cdot$ to denote the sesquilinear form and $|\cdot|$ to denote the norm. This is mostly a matter of taste, but depending on the vector space $V$ the notations $|\cdot|$ and $\cdot$ may already be occupied. Take for example the real vector space $C([0,1])$ of the continuous realvalued functions on the unit interval, together with the scalar product $$ \langle f,g \rangle = \int_0^1 f(t) g(t) \,\text{d}t. $$ Then the notation $f \cdot g$ is already occupied with the pointwise product of $f$ and $g$ and $|f|$ with the pointwise absolute value of $f$.
Regarding the left side of the inequality the first thing to realize is that in the case of a complex vector space $V$ the scalar $\langle v, w \rangle$ is a complex number and not necessarily real. But the inequality $$ \langle v, w \rangle \leq \|v\| \cdot \|w\| $$ does not make sense if $\langle v, w \rangle$ is not real. So in this case one can not omit the absolute value.
If we are working over the real numbers, then it is possible to omit the absolute value from the left side of the inequality without weaking the inequality. To see this notice that $x \leq |x|$ for all $x \in \mathbb{R}$, so it follows from the Cauchy-Schwarz inequality that $$ \langle v, w \rangle \leq |\langle v, w \rangle| \leq \|v\| \cdot \|w\| \quad \text{for all $v,w \in V$}. $$ Interestingly the reverse also follows, because for every $v,w \in V$ we then have \begin{gather*} \langle v, w \rangle \leq \|v\| \cdot \|w\| \quad\text{and}\quad -\langle v, w \rangle = \langle -v, w \rangle \leq \|-v\| \cdot \|w\| = \|v\| \cdot \|w\|, \end{gather*} and thus $ |\langle v, w \rangle| = \max\{\langle v,w \rangle, -\langle v, w \rangle\} \leq \|v\| \cdot \|w\|. $