I think we established that the literature is lacking on this question. But I think the "correct" definition of morphisms between hyperstonean spaces can be puzzled together from G. Bezhanishvili's paper "Stone duality and Gleason covers through de Vries duality" (Topology and its Applications 157:1064-1080, 2010), especially section 6.
He proves in detail a duality between the category of complete Boolean algebras and complete Boolean algebra homomorphisms, and the category of extremally disconnected compact Hausdorff spaces and continuous open maps. But commutative von Neumann algebras and normal *-homomorphisms form a full subcategory of the former (via taking projections), which corresponds to the full subcategory of the latter consisting of hyperstonean spaces.
So Gelfand duality really restricts quite cleanly: commutative von Neumann algebras and normal *-homomorphisms are dual to hyperstonean spaces and open continuous maps.
The general setup in topological spaces would be the following: given topological spaces $X,Y$ and a dense $X_0\subseteq X$ with $T:X_0\rightarrow Y$ continuous (for the subspace topology on $X_0$) we can extend $T$ by continuity to $X$. This is of course not true (exercise to the reader to find a counter-example for e.g. $X_0=(0,1), X=[0,1]$).
A true statement can be fashioned by using the theory of Uniform Spaces, Uniform Continuity and Completeness. I believe that then a general statement could be formulated and proved for topological vector spaces.
However, in this case, I think there are some missing details in Takesaki's proof, and I would prefer to give a different approach.
The way I would think about this problem is to use some Banach space theory. The general setup is to let $E,F$ be Banach spaces, let $E_0\subseteq E^*$ be a weak$^*$-dense subspace, and let $T_0:E_0\rightarrow F^*$ be bounded, linear, and weak$^*$-continuous. We claim we can extend $T_0$ to $T:E^*\rightarrow F^*$ a weak$^*$-continuous bounded linear map.
In fact, Takesaki proves something different. We can consider the Banach space adjoint (here I follow Takesaki's notation) ${}^t T_0:F^{**}\rightarrow E_0^*$, consider $F$ as a subspace of $F^{**}$ and so ask about ${}^t T_0(F) \subseteq E_0^*$.
Also each member of $E$ induces a functional on $E_0$, and by Hahn-Banach, weak$^*$-density of $E_0$ in $E^*$ implies that the resulting map $E\rightarrow E_0^*$ is at least injective. Takesaki shows that ${}^t T_0(F) \subseteq E$. So there is a linear map $S:F\rightarrow E$ making the following commute:
$$ \require{AMScd}\begin{CD} F^{**} @>{{}^t T_0}>> E_0^* \\
@AAA @AAA \\
F @>{S}>> E \end{CD} $$
Explicitly, given $\omega\in F$ and $x\in E_0\subseteq E^*$,
$$ \langle x, S(\omega) \rangle_{E^*,E} =
\langle {}^t T_0(\omega), x \rangle_{E_0^*, E_0}
= \langle T_0(x), \omega \rangle_{F^*,F}. $$
In general, I see no reason why $S$ need be bounded. Kaplansky density comes to the rescue, because $M_1 \otimes_{\min} M_2$ is a weak$^*$-dense $C^*$-subalgebra of $M_1 \overline\otimes M_2$, and so the unit ball is dense. Abstractly, this means we may add the additional hypothesis that the unit ball of $E_0$ is weak$^*$-dense in the unit ball of $E^*$. Equivalently (by Hahn-Banach again) $E_0$ norms $E$. So
\begin{align*}
\|S(\omega)\| &= \sup \{ |\langle x,S(\omega)\rangle| : x\in E_0, \|x\|\leq 1 \} \\
&= \sup \{ |\langle T_0(x), \omega \rangle| : x\in E_0, \|x\|\leq 1 \} \\
&\leq \|T_0\| \|\omega\|.
\end{align*}
Thus $S$ is bounded, with $\|S\| \leq \|T_0\|$.
Now set $T = {}^t S_0 : E^*\rightarrow F^*$ and it's an easy exercise to show that $T$ extends $T_0$; clearly $T$ is weak$^*$-continuous as it's a Banach space adjoint.
Best Answer
"Yes".
Degenerate representations of $C^*$-algebras have a simple form: each $\pi(a)$ restricts to $H_1 := [\pi(A)(H)]$ and acts as $0$ on $H_1^\perp$. Thus we obtain a non-degenerate representation $\pi_1$ on $H_1$, and the $0$ representation on $H_1^\perp$.
A calculation shows that $\tilde\pi:A^{**}\rightarrow B(H)$ will have the same decomposition. However, $\pi(A)''$ will be $\{ T+\alpha 1_{H_1^\perp} : T\in \pi_1(A)'', \alpha\in\mathbb C\}$, compare with Theorem 3.9 in the previous chapter of Takesaki. So we can never hope to have that $\tilde\pi$ maps onto $\mathcal M(\pi) = \pi(A)''$, unless $\pi$ is non-degenerate.
I think there are some other statements in this section of Takesaki which need "non-degenerate". However, we've just seen that degenerate representations are in a sense trivial, and so really we can just read "representation" as always meaning "non-degenerate representation".
(Notice there is another small typo: surely $\mathcal M$ should be $\mathcal M(\pi)$ in the statement of (iii). This is just a fact of life: authors are human, and we have to expect the occasional slip.)