Turing Machine Decidable: What right does the definition have to say what's not in language L?
(I'll convert the comment into an answer)

First, definitions:

$L$ is a language, that is, a set of words where each word has a finite length. If all the words that ever exist are denoted as $\Sigma^*$ then any language is $L\subseteq \Sigma^*$. Words that are not in $L$ are in a different subset we denote $\overline L=\Sigma^*\setminus L$. Both $L$ and $\overline L$ can be infinite (i.e., have infinite many words). Still, each word in $L$ or in$\overline L$ is of finite length.

Now, assume that we have a TM that never goes into an infinite loop (such TMs are called deciders). If it never goes into a loop, it means that for any input word, it will either halt and accept or halt and reject.

If we used your second definition, then any $L$ will be $L=\Sigma^*$, since all the words are either accepted or rejected.

So the answer is that the first definition is correct. A decider $M$ (a TM that always halt!) splits the space of all the possible words $\Sigma^*$ into two disjoint subsets, the words that are accepted by $M$, denoted $L(M)$ and the ones rejected by $M$, denoted $\overline{L(M)}$. Again, $\Sigma^*=L(M) \cup \overline{L(M)}$, since every word is either accepted or rejected.