Why if L and ¬L (Complement of L) are turing-recognized both are turing-decidable?
I will try to explain it in simpler terms.
If $L$ is Turing recognizable, there exists a TM $M_1$ that halts and accepts any input string $w \in L$. It is not guaranteed to halt on any input string $w \notin L$.
If $\bar{L}$ is Turing recognizable, there exists a TM $M_2$ that halts and accepts any input string $w \notin L$. Similarly to above, $M_2$ is not guaranteed to halt on any input string $w \in L$.
Now, for every $w$ we have two options: either $w \in L$ or $w \notin L$. Therefore for any input string $w$, exactly one of these machines will halt and accept. This allows us to construct the decider described in you proof. Create a machine $M$ that simulates $M_1$ and $M_2$. With $w$ as input, accept $w$ if $M_1$ accepts, reject if $M_2$ accepts. This machine is a decider for $L$ and therefore $L$ is decidable.
To prove $\bar{L}$ decidable as well, you could apply the same logic, but it is sufficient to point out that decidable languages are closed under complement.