Complexity of a Turing Machine when changing its alphabet to binary
The first part the proof reads "To simulate one step of $M$, the machine $M^{\sim}$ will: (1) ... (2)... (3)... (4) and (5) ....". Can you see the factor of 4?

You have a good understanding regarding the factor $\log|A|$.  Let us take a close look at the steps in each stage of $M^{\sim}$ above. There are $\log|A|$ steps in stage (1) and in stage (5). However, there is only one step in stage (2) or stage (3) or stage (4). So there are less than $2\log|A|$ steps in stage (2), (3) and (4) combined, assuming $\log|A|\gt1$. If $\log|A|=1$, we can just let $M^\sim$ be $M$ and we are done. So there are less than $\log|A|+\log|A|+2\log|A|=4\log|A|$ steps of $M^\sim$ for one step of $M$. 

The second part of the proof is justifying that the stage (2), (3) or (4)  can be done in one step of machine $M^{\sim}$. It explains the power of larger registers. It points out that we can use more states in $M^\sim$ to identify the possibly much more situations ("states" could have been a better word had we not reserved that word for Turing machine) of the machine. Take your time. You will understand.