How come this turing machine running time is $O(n^{1+\log_2 2006}\cdot \log_2n)$
This bound comes from the number of possible configurations. If you do more steps than you have configurations, you are in a cycle.

So let us quickly get a bound for the number of configurations. You have $100+\log n$ possible positions for the head on the work tape and $n$ positions for the head on the input tape. The work tape can contain $2006^{100+\log n}$ different words. Furthermore there are 27 states. Combining everything gives 
$$
\begin{align}
27n \cdot (100+\log n) \cdot 2006^{100+\log n}&amp;=27\cdot2006^{100}(100n+n\log n) n^{\log 2006}\\
&amp;=O((100n+ n\log n)  n^{\log 2006})\\
&amp;=O(\log n\cdot  n^{1+\log 2006})\\
\end{align}$$