This is actually much simpler than you may suspect: $\omega_\alpha^{CK}$ is well-defined for **every** ordinal $\alpha$, not just the countable ones, if we use the set-theoretic as opposed to computability-theoretic definition. Specifically, we define $\omega_\alpha^{CK}$ as the unique ordinal $\eta$ such that

 - $L_\eta\models$ KP (that is, $\eta$ is **admissible**)

and

 - $\{\gamma<\eta: L_\gamma\models\mbox{ KP}\}$ has ordertype $\alpha$.

That is, $\omega_\alpha^{CK}$ is the $\alpha$th admissible ordinal. The agreement with the computability-theoretic definition at countable levels is a theorem of Sacks: specifically, a countable ordinal is admissible iff it is the least ordinal with no $r$-computable copy for some real $r$.

For example, it's easy to check that in fact $$\omega_1=\omega^{CK}_{\omega_1}$$ that is, $\omega_1$ is a fixed point of the "admissible-counting" function. (It's definitely not the *least* fixed point, of course.)

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Via forcing, we can give a computability-theoretic interpretation of $\omega_\alpha^{CK}$ even when $\alpha$ is uncountable *(CAVEAT: this is my own work, so you should take my approval of it with a grain of salt)*:

 - Say that an ordinal $\gamma$ is **generically Church-Kleene** if in some generic extension of the universe in which $\gamma$ is countable, there is some real $r$ such that $\gamma$ is the least ordinal with no $r$-computable copy.

  - By [Shoenfield's absoluteness theorem](), we can replace "some generic extension" with "every generic extension" above; in particular, this means that this agrees with the usual notion when $\gamma$ is countable.

 - Then we can prove - by "genericizing" Sacks' theorem - that $\omega^{CK}_\alpha$ is exactly the $\alpha$th generically Church-Kleene ordinal; that is, $\omega_\alpha^{CK}$ is the unique generically Church-Kleene ordinal such that the set of smaller generically Church-Kleene ordinals has ordertype $\alpha$.

*It's worth contrasting this with the perspective given by admissible recursion theory: in general, the supremum of the $\alpha$-recursive well-orderings of $\alpha$ is [**vastly smaller than**](https://mathoverflow.net/questions/277343/order-type-of-alpha-computable-well-orderings/277345#277345) the next admissible above $\alpha$. This is true, for example, when $\alpha=\omega_1$. Ordinals where this doesn't happen are called [Gandy ordinals](https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/jlms/s2-14.3.387), and in a precise sense, most ordinals - even most countable ordinals - are not Gandy.*

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Re: your second question, I'm not too familiar with infinitary Turing machines, but I'm fairly certain they all get you well past the next admissible; so the answer should be **yes**.