As stated, $M'$ is a well defined function. Sometimes $M'$ is computable, as in simple cases such as when $M$ is a machine that always halts or never halts. Is $M'$ always computable? It turns out not.
Consider the case of a semidecidable set $\Gamma$. Let $M$ be a Turring machine that outputs $1$ for all $\alpha\in\Gamma$, and outputs $0$ or loops for $\alpha\notin\Gamma$. First, note that $M$ is not a universal Turring machine. Proof: If $M$ is universal, then $M(\alpha)$ simulates $\alpha(0)$. Then we have $\alpha\in\Gamma$ iff $\alpha(0)=1$. Therefore all $\alpha\in\Gamma$ output $1$ when run on a blank tape. $\Gamma$ was an arbitrary semidecidable set, so the result holds as well for all semidecidable sets. However, the set of all Turing machines is decidable, and thus, semidecidable. But there are some Turring machines that do not output $1$ when run on a blank tape. Therefore, $M$ is not universal.
Since $M$ is not universal, we can define $M'$. Assume $M'$ is computable (to show a contradiction). Now $M'$ outputs $1$ for all $\alpha\in\Gamma$ and outputs $0$ otherwise. Therefore, $\Gamma$ is decidable. That means that all semidecidable sets are decidable. A contradiction, and thus, $M'$ is uncomputable.
Finally, your question is whether we can define a machine (as opposed to a function) $M'$ for all non-universal $M$. As shown above, we cannot define such a machine when $M$ is a machine that enumerates a semidecidable set.
In general, for what $M$ can we define (the machine) $M'$, and when can't we? A machine $M'$ exists iff the set $G$ of all values for which $M$ halts is decidable. Proof: Suppose the machine $M'$ exists. $M(\alpha)$ halts iff $M'(\alpha)=1$, and $M(\alpha)$ does not halt iff $M'(\alpha)=0$, so $G$ is decidable. Suppose $G$ is decidable. Then there is a machine that accepts $\alpha$ if $M(\alpha)$ halts, and rejects $\alpha$ if $M(\alpha)$ does not halt. That is machine $M'$.