How is $\varepsilon_+^\mu(p) = \bar{v}(k) \gamma^\mu u(p)$ derived? The relation $$\varepsilon_+^\mu(p) = \bar{v}(k) \gamma^\mu u(p)$$ is sometimes used to ease calculations of Feynman amplitudes with external gluons (see for example here at (2.13)). Where does this relation come from?
How can we use the above relation to show that
$$\varepsilon_+(p) \cdot \varepsilon_+(q) = 0 \,\, \forall p,q,$$
and can this last equation also be obtained throught some gauge transformation, without using the relation between polarization vectors and Dirac spinors?
 A: You have stumbled upon the spinor helicity formalism. The idea is as follows. Massless spinors have many useful relationships among them that we can use. Thus instead of using the polarization vector, $\epsilon_\mu$ or massive spinors lets just map these objects into massless spinors and then use the convenient relationships between them to simplify amplitudes. 
In this particular case you are looking at a mapping for the polarization vector. By Lorentz invariance the expression, $A\bar{v} \gamma^\mu u $ is the simplest thing you can write down (I guess you could try a derivative instead of $\gamma_\mu$ but this might lead to problems down the road). 
Since $\epsilon_\mu$ represents a massless gauge field it has 2 degrees of freedom, while $\bar{v}\gamma^\mu u $ has in 4 degrees of freedom. Therefore,  $v$ must not be a physical quantity. It just needs to be fixed when choosing a gauge.
With regards to your identity we have,
\begin{align}
\epsilon _\mu ^+ \cdot \epsilon ^{+ \mu } & = A ^2 \left\langle  p + | \gamma _\mu | k + \right\rangle  \left\langle  p + | \gamma ^\mu | k + \right\rangle  \\ 
& = A ^2 \left\langle  p + | \gamma _\mu | k + \right\rangle  \left\langle  k - | \gamma ^\mu p - \right\rangle  \\ 
& = 2A ^2 \left\langle  p + | p - \right\rangle  \left\langle  k - | k + \right\rangle \\
& = 0
\end{align} 
where we have used the massless spinor relations, $ \left\langle  A + | \gamma ^\mu | B + \right\rangle  = \left\langle  B - | \gamma ^\mu A - \right\rangle  $ and $ \left\langle  A + | A - \right\rangle  = 0 $. These are both easy to derive by returning to $u, \bar{u}$ notation and manipulating the spinors.  
