One of my exercises was to prove the identity $$\gamma^\mu\displaystyle{\not}a\gamma_\mu=-2\displaystyle{\not}a.$$
Which is trivial if $\gamma^\mu a_\nu=a_\nu \gamma^\mu$, as follows $$\gamma^\mu\displaystyle{\not}a\gamma_\mu=\gamma^\mu a_\nu \gamma^\nu\gamma_\mu = -2 a_\nu \gamma^\nu=-2\displaystyle{\not}a.$$
Yes, they do. Something complicated about quantum field theory is that we often hide the countless tensor products that are in effect. If we use $a,b$ as spinor indices (i.e., indices labeling the spinorial entries of spinorial matrices and spinors), we can write \begin{align} (\gamma^\mu \displaystyle{\not}a \gamma_\mu)_{ab} &= \gamma_{ac}^\mu \displaystyle{\not}a_{cd} \gamma_{\mu,db}, \\ &= \gamma_{ac}^\mu a_{\nu} \gamma^{\nu}_{cd} \gamma_{\mu,db}. \end{align} This time, all terms are numbers, since we are working in components. Thus,
\begin{align} (\gamma^\mu \displaystyle{\not}a \gamma_\mu)_{ab} &= \gamma_{ac}^\mu a_{\nu} \gamma^{\nu}_{cd} \gamma_{\mu,db}, \\ &= a_{\nu} \gamma_{ac}^\mu \gamma^{\nu}_{cd} \gamma_{\mu,db}, \\ &= a_{\nu} (\gamma^\mu \gamma^{\nu} \gamma_{\mu})_{ab}, \end{align} at which point you can just follow your calculation and things run smoothly.
The key point is noticing that four-vector do not carry spinor indices.
