The reason is that the definitions of ordinal arithmetics and cardinal arithmetics are very different.
Whereas the ordinal arithmetics operations are concerned with order types, the cardinal arithmetics are concerned with certain sets.
For example, $\alpha+\beta$ as ordinals is the order type of $\alpha$ concatenated with $\beta$. Whereas cardinality strips naked any possible structure, and considering the cardinality of the set $\{0\}\times\alpha\cup\{1\}\times\beta$, which is equal to the maximum of $|\alpha|$ and $|\beta|$ (granted one is infinite).
Exponentiation, which is the strangest one, is defined very differently, again, from ordinals and cardinals.
- In cardinals $\alpha^\beta$ is the cardinality of the set of all functions from $\beta$ to $\alpha$.
In the ordinals, we care about the order, so $\alpha^\beta$ is the order type of the reverse lexicographic order of functions from $\beta$ into $\alpha$ which are non-zero only in finitely many coordinates.
Equivalently, and perhaps more clearly, we can define this by induction, $\alpha^0=1$, $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$, and for a limit ordinal $\delta$, $\alpha^\delta=\sup\{\alpha^\beta\mid\beta<\delta\}$.
Now we easily see that $2^\omega=\omega$ when we are talking about ordinal exponentiation. $2^\omega$ is the limit of $2^n$ for finite $n$, but $2^n$ has a finite order type, and it's a strictly increasing sequence. The limit of a strictly increasing sequence of finite ordinals is $\omega$ itself.
Well, it sounds weird, isn't it? But it's not really weird. We have this sort of phenomenon in other - more familiar - systems of arithmetics.
In the natural numbers $n\cdot m$ can be defined as repeated addition of $n$, $m$ times. Addition itself can be defined as repeated successor operation. On the other hand, when we consider the real numbers $\sqrt2\cdot\sqrt2$ cannot be thought as repeated addition. What does it even mean to add something $\sqrt2$ times? It is true that in this case, if we restrict back to the natural numbers then the operations become repeated application of the "previous operation", but this is because of the nature of the natural numbers as a corner stone of modern mathematics (in many many ways). For infinite, and less-corner stoney this is not the case, as exhibited in ordinals and cardinals.
Both questions are answered in the paper you refered to (i.e., "Recognizable sets and Woodin cardinals: computation beyond the constructible universe") https://arxiv.org/abs/1512.06101 in Lemma 3.13 (I will write "OTM-aw" and "OTM-ew" for "OTM-accidentally writable" and "OTM-eventually writable":
Concerning accidental writability: You are right, the OTM-aw real numbers are exactly the constructible ones. More or less, one can see this in the way you indicated by simultaneously simulating all OTM-programs in all parameters and, whenever one of these outputs a real number, one writes this to the output section.
Concerning eventual writability: If $\eta$ is minimal such that $L_{\eta}$ is a $\Sigma_{2}$-submodel of $L$ then $x\subseteq\omega$ is OTM-ew if and only if $x\in L_{\eta}$.
However, it is possible, as you indicate, to see that the supremum of the OTM-ew ordinals must be countable without determining its value: Each coded ordinal will be countable in $L$, and there are only countable many programs, hence at most countably many OTM-ew ordinals. Finally, the function $f:\omega\rightarrow\omega_{1}^{L}$ mapping $i$ to the ordinal eventually written by the $i$th OTM-program (if it exists) and to $0$, otherwise, is definable in $L$, hence contained in $L$, and now the supremum of the OTM-ew ordinals is $\bigcup f[\omega]$, which is countable in $L$.
I understand that part B is asking what is wrong with the following method to eventually write the supremum of the OTM-ew ordinals: Run all program simultaneously, write a code for the sum of the outputs to the output section. At some point, all programs that stabilize have stabilized and then, the output will be stable and equal the sum of all OTM-ew ordinals.
The problem is that, if we do this with all programs, then the output will not stabilize, because it will include the outputs of non-stabilizing programs. It would work if we could restrict ourselves to those programs that stabilize; however, the set of programs that stabilize is quite complicated and in particular not OTM-ew (basically, this section is a reductio proof of this), so this also does not work.
edited: Previously, this post stated that $\eta$ merely needs to be minimal such that $L_{\eta}$ and $L$ have the same $\Sigma_{2}$-theory, which would mean that it is an $L$-countable ordinal. As Joel Hamkins pointed out below, this is false.
Best Answer
If you read the sentence "just before" definition-3.10 then you will see the expression $x \subseteq \gamma$. So if $\gamma$ is some ordinal then it seems to me that the author(s) are talking about some subset $x$ of the $\gamma$. That is a set of ordinals whose where each individual ordinal must be less than $\gamma$. So to define "accidentally writeable reals" specifically, you just need to set $\gamma=\omega$ (in definition-3.10 for the paper you linked).
Further regarding this point:
It seems that no new accidentally writeable real can be generated at some time $\geq \omega_1$. That's because if it was possible, then given some specific real number as input to OTM, it would be possible to halt at some time $\geq \omega_1$, which is impossible [ for this, start enumerating all accidentally writeable reals and when a new accidentally written real matches the "input real" then halt ]. For this, I don't know the reason, but I think that the answer to one of your older questions is relevant in this regard https://mathoverflow.net/questions/372051. I am assuming that the answer given is correct (since I don't understand it).
Edit: Sorry I misremembered the reference I was talking about.
Regarding this point:
This should be equal to $\omega^L_1$. It is consistent with set theory that $\omega^L_1=\omega_1$. It is also consistent that $\omega^L_1 \neq \omega_1$. I don't know the reason, but the answer in this question seems to mention this: Existence of bijections in $L$.