The given operation is done with $6$ significant digits (mantissa) and in base $10$
$$10^6+\underbrace{1+1+\ldots+1}_{10^6\text{-times}} $$
Now we had to get $2*10^6$ but we get $10^6$, why is that?
I understand that the number is stored as:
$\sigma\cdot (m)_{B}\cdot B^{exp}$
So $$10^6=(-1)^0*1.000000*10^6$$
So $$1=(-1)^0*1.000000*10^0\text{ or }1=(-1)^0*0.000001*10^6?$$ Why does it turn out that $10^6+1=10^6?$
Perhaps because to $6$ significant digits, $\color{#C00}{100000}1=\color{#C00}{100000}0$ when rounded. If you are really working in base ten, unlike binary, you can't assume the first digit is $1$ and only count the remaining digits for the $6$ significant digits. This would mean that only the parts in red are being retained by the calculator/computer.
Try incrementing by $10$ and see if the additions stick.