Dot product of vector and its derivative with respect to time? How does $L \cdot\frac{dL}{dt} = \frac{1}{2}\frac{d(L^2)}{dt}$? How does:
$$L \cdot\frac{dL}{dt} = \frac{1}{2}\frac{d(L^2)}{dt}$$
where L is a vector (I dunno how to make it bold in the equation). 
How do they reach to this right hand side equation?
And what is the difference between ||L|| and |L| ?
 A: From math and the power rule: $\dfrac{d(x^2)}{dx} = 2x$
And we assume that L is a function of time: $\vec{L} = \vec{L(t)}$.
To refresh you on the chain rule: if x were a function of time, then $\dfrac{d(f(x))}{dt} = \dfrac{d(f(x))}{dx} * \dfrac{dx}{dt}$.
Back to math: $\dfrac{d(x^2)}{dx}$ is actually $2x\dfrac{dx}{dx}$ if you apply said chain rule.  $\dfrac{dx}{dx}$ is commonly held to be equal to $1$, so it's left out.
As such: $\dfrac{d(\vec{L}^2)}{dt} = 2\vec{L}\dfrac{d\vec{L}}{dt}$.
This is all assuming that we're operating element-wise on your vector $\vec{L}$.  That means it's the same as a normal (scalar) equation, but there is one scalar equation for each dimension of your vector.
Additionally, the notation $\vec{L}^2$ for a vector should be avoided.  Use dot product or cross product.  This equation should be written as:
$$2\vec{L}\cdot\dfrac{d\vec{L}}{dt} = \dfrac{d(\vec{L}\cdot\vec{L})}{dt}$$
This equation is not true if $L^2$ were to be interpreted as a cross product ($\vec{L}\times \vec{L} = 0$) of a vector with itself.
As for the difference between single and double bars, both mean magnitude, but for vectors we like to use double bars to remove any similarity to the absolute value of a number:
$$|-3| = 3$$
$$||\vec{\langle 3, 4\rangle}|| = 5$$
A: It is the easiest to think of this problem in the component form. Let's say the vector $\vec L$ is in a 3 dimensional space (that is usually what we use). So in component form, i.e., written as a 3 by 1 matrix, $\vec L=(L_1,L_2,L_3)$. 
What is the left-hand-side of your expression? It is 
$\vec L\cdot\frac{\mathrm d\vec L}{\mathrm dt}=L_1\cdot\frac{\mathrm dL_1}{\mathrm dt}+L_2\cdot\frac{\mathrm dL_2}{\mathrm dt}+L_3\cdot\frac{\mathrm dL_3}{\mathrm dt}$
Now, you can consider each term on the right-hand-side. In fact, we only need consider the first term, which, as you can remember,
$L_1\cdot\frac{\mathrm dL_1}{\mathrm dt}=\frac{1}{2}\frac{\mathrm dL^2_1}{\mathrm dt}$
Because of the linearity of derivative, we finally arrive at 
$\vec L\cdot\frac{\mathrm d\vec L}{\mathrm dt}=\frac{1}{2}\frac{\mathrm dL^2_1}{\mathrm dt}+\frac{1}{2}\frac{\mathrm dL^2_2}{\mathrm dt}+\frac{1}{2}\frac{\mathrm dL^2_3}{\mathrm dt}=\frac{1}{2}\frac{\mathrm d}{\mathrm dt}(L^2_1+L^2_2+L^2_3)=\frac{1}{2}\frac{\mathrm d\vec L^2}{\mathrm dt}$
A: This is true for any vector quantity from a finite-dimensioned vector space that uses the standard definition of the inner product. Let $\mathbf u$ and $\mathbf v$ be elements of the vector space $\mathbb R^N$ with inner product $\mathbf u \cdot \mathbf v = \sum_{i=1}^N u_i v_i$. Then $\mathbf u \cdot \mathbf u = \sum_{i=1}^N {u_i}^2$. Using the standard notation $u^2 = \mathbf u \cdot \mathbf u$ and differentiating with respect to time yields
$$\begin{aligned}
\frac {d\,u^2}{dt} &= \frac {d}{dt} \bigr(\mathbf u(t) \cdot \mathbf u(t)\bigr) && \text{step 1, see note #1}\\
&= \frac {d}{dt} \left(\sum_{i=1}^N u_i(t) u_i(t)\right) && \text{step 2, see note #2}\\
&= \sum_{i=1}^N \frac {d}{dt} \bigl(u_i(t) u_i(t)\bigr) && \text{step 3}\\
&= \sum_{i=1}^N u_i(t)\dot u_i(t) + \dot u_i(t) u_i(t) && \text{step 4}\\
&= 2\sum_{i=1}^N u_i(t)\dot u_i(t) && \text{step 5, see note #3}\\
&= 2 \mathbf u \cdot \dot{\mathbf u} && \text{step 6, see note #1}
\end{aligned}$$
Note well that I used a number of "tricks" here:


*

*The transitions from step #1 to step #2 and from step #5 to step #6 assume the standard Euclidean definition of the inner product. There are lots of other inner products out there!

*I switched the sum and derivative between steps #2 and #3. That's valid if the space is of finite dimension. It's not necessarily valid in the case of an infinite dimensional vector.

*In going from step #4 to step #5, I assumed that $u_i(t)\dot u_i(t) + \dot u_i(t) u_i(t) = 2u_i(t)\dot u_i(t)$. That's only valid if multiplication is commutative. While that certainly is the case for real and complex functions, it is not true in general.
A: If $L(t)$ were a scalar valued function of time then, by the product rule, we have
$$\frac{d}{dt}L^2(t) = \frac{d}{dt}\left(L(t) \cdot L(t) \right) = \frac{dL}{dt} \cdot L(t)+ L(t) \cdot\frac{dL}{dt} = 2L(t)\cdot\frac{dL}{dt}$$
Due to linearity, this holds for ordinary vector valued functions of time since
$$L^2 = \vec L \cdot \vec L$$
Thus
$$\frac{d}{dt}L^2(t) = \frac{d}{dt}\left(\vec L(t) \cdot \vec L(t) \right) = \frac{d\vec L}{dt} \cdot \vec L(t)+ \vec L(t) \cdot\frac{d\vec L}{dt} = 2\vec L(t)\cdot\frac{d\vec L}{dt}$$
