First of all I should mention that "Finite Registers" could perhaps more appropriately be written as "Finite Number of Registers". I just tried to shorten the phrase to avoid the title from becoming too long.
The question is a fairly limited/specific version of a more general question I had in mind. However, I felt that perhaps it might be more suitable to ask the limited question first.
Suppose we are given some computable well-ordering on $\Bbb{N}$ whose order-type is $\omega$. It is assumed that the (total) notation function induced corresponding to the computable less-than relation is denoted as $address:\omega\rightarrow \Bbb{N}$ and assumed to be bijective.
Now suppose we are given two registers (generalising to more than two registers is self-evident). The values of registers are given by the functions $Value_1:\omega\rightarrow\omega$ and $Value_2:\omega\rightarrow\omega$. More specifically, the value of register-1 at some time $t\in\omega$ is denoted by $Value_1(t)$ (and of course same for the other register).
Before stating this question we need to define the representation of a function in a given computable well-ordering. Now suppose for some computable well-ordering of an element $p$ we are given the address function $address:p\rightarrow \Bbb{N}$. For some function $F:p \rightarrow p$ we can write representation function of $F$ (in the given well-ordering for $p$) as $f:\Bbb{N} \rightarrow \Bbb{N}$ and defined by: $$f(address(x))=address(F(x)) \qquad for \; all \; x\in p$$
In general we might have to worry about extra cases (in defining representation of a function), but here we won't (since $F$ is total and $F(x)<p$ for all values in the domain). One can also observe that since we are only concerned with case of $\omega$ here, we can replace $p$ with $\omega$ in the preceding paragraph.
And with this, we are pretty much set to ask the question. We just need to define the rules for how the register values can change. They are:
(1) We must have: $$Value_1(0)=0$$ $$Value_1(1)=0$$ And similarly for second register.
(2) If we have: $$Value_1(t)=a$$
Then only one of the following possibilities can hold at $t+1$:
(a) $Value_1(t+1)=a$ (value remains unchanged)
(b) $Value_1(t+1)=a+1$ (value increased by $1$)
(c) $Value_1(t+1)=0$ (value brought down to $0$)
And obviously the same rule holds for the second register.
(3) There must not exist two different times $t_1$ and $t_2$ (with $t_1\ne 0$ and $t_2\ne 0$) such that: $$Value_1(t_1)=Value_1(t_2)$$ AND $$Value_2(t_1)=Value_2(t_2)$$
Now the question asks that can one find a computable well-ordering for $\omega$ such that:
(1) The representation functions corresponding to the functions $Value_1$ and $Value_2$ are recursive in the well-ordering.
(2) The representation function corresponding to the successor function is not recursive in the well-ordering.
In the case of positive answer an example can be given. In case of negative answer the proof can be given (also generalisation to more than two resgisters would be nice of course).
P.S. Few readers might have noted that this is much more limited version of a question I posted on MO (that wasn't received well). Unfortunately, on top of question being fairly long, I made a "question breaking" mistake there (which I didn't realise at the time of posting).
edit: made some minor edits (changed a few words here and there to clarify the meaning further)