With regard to the deeper motivation behind this question: I think there's some confusions around exactly how to interpret nonstandard models. I'm going to focus on the very narrow question, "Can models disagree about the computability of certain definable sets?"
The first step is realizing that this is a bit vague. There's at least three ways I can see to interpret it:
Interpretation 1 (Internal). You give me a formula in the language of arithmetic - viewed as a definition of a set - $\varphi(x)$. Each model $M$ of $PA$ yields a set $\varphi^M$. Now, $\varphi^M$ is a subset of $M$, not $\mathbb{N}$, but $M$ can't tell the difference. The sentence $\psi\equiv$"$\{x: \varphi(x)\}$ is decidable(=computable)" is expressible in the language of arithmetic, as $$\psi\equiv\exists e_0, e_1\forall x[(\varphi(x)\iff \exists sT(e_0, x, s))\wedge (\neg\varphi(x)\iff\exists sT(e_1, x, s))],$$ where $T$ is Kleene's $T$ predicate: $T(e, x, s)$ means "the $e$th Turing machine halts on input $x$ in $s$ steps."
Okay, fix $\varphi$. We can now ask:
Is there a model $M$ which satisfies $\psi$? $\neg \psi$?
Interpretation 2 (External). You give me a formula in the language of arithmetic - viewed as a definition of a set - $\varphi(x)$. Each model $M$ of $PA$ yields a set $\varphi^M$. Now, $\varphi^M$ is a subset of $M$, but the standard model $\mathbb{N}$ injects into $M$ in a canonical way; call this map "$i_M$." So we can talk about the standard natural numbers which ($M$ thinks) the statement $\varphi$ is true of; that is, the set $$X_M=i_M^{-1}(\varphi^M).$$ This set $X_M$ is a genuine set of natural numbers, and we can ask whether $X_M$ is: computable, infinite, coinfinite, of asymptotic density 1, etc. For instance, in the Goodstein example, $X_M$ is the set of standard natural numbers which $M$ thinks have halting Goodstein sequences.
This "cutting off" process might seem unnatural, but it's actually really mathematically interesting: it's what standard systems, for instance, are all about.
Okay, fix $\varphi$. We can now ask:
Is there a model $M$ such that $X_M$ is computable? Non-computable?
Interpretation 3 (Mix-and-match). As in interpretation 1, we look at the whole set $\varphi^M$. However, we now look at standard indices for computable functions, interpreted in $M$. Specifically, say $\varphi^M$ is mix-and-match-computable in $M$ if there is there some standard natural number $n$ such that $M$ thinks $\Phi_n$ is a total computable function such that $\Phi_n$ is the characteristic function of $\varphi^M$.
Okay fix $\varphi$. We can now ask:
Is there a model $M$ such that $\varphi^M$ is mix-and-match computable in $M$? Non-mix-and-match computable in $M$?
Despite the heterogeneous nature of this question, I suspect it too has its place in your intuition: basically, this question is asking, "Is there a genuine algorithm which, consistently, solves this problem?"
Now, what about your actual question? Well, we can exhibit formulas $\varphi$ such that there are models exhibiting divergent behavior in each of the three interpretations. This is actually pretty easy to do: we just shoehorn in some known undecidability. For instance, consider the formula $$\varphi(x)\equiv [\neg Con(PA) \wedge x\not=x]\vee[Con(PA) \wedge \exists s(T(x, x, s)].$$ In a model of $PA$ which thinks $PA$ is inconsistent, $\varphi$ just denotes the empty set, but in any model of $PA$ which thinks $PA$ is consistent - such as the standard one - $\varphi$ denotes that model's halting problem, which that model will think is incomputable. This might seem a bit trivial - it certainly does to me - but there's not an obvious way I can see to "carve out" such trivial solutions. (Maybe someone here can think of one!)
First you ask
how Cantor make it sure that bijection of two infinite sets can also
ensure the two infinite sets have the same size?
The answer, which you seem to understand, is that he didn't "make it sure", he made that the definition of "have the same size".
Then you note, correctly, that this definition leads to many counterintuitive results, adding
I think if we find another standard to compare the size of two
infinite sets and also make the result useful and accord with our
intuition, that would be much better .
Perhaps. But mathematicians haven't succeeded in finding another definition that's as generally useful. They've taken an alternative path, refining their intuition about "same size" so as to give results they can prove from Cantor's definition.
You might want to read about some of the controversy surrounding Cantor's work. Some mathematicians at the time tried to do what you suggest. David Hilbert championed Cantor:
No one shall expel us from the Paradise that Cantor has created. (https://en.wikiquote.org/wiki/David_Hilbert)
Cantor's paradise is an expression used by David Hilbert (1926, page
170) in describing set theory and infinite cardinal numbers developed
by Georg Cantor. The context of Hilbert's comment was his opposition
to what he saw as L. E. J. Brouwer's reductive attempts to
circumscribe what kind of mathematics is acceptable; see
Brouwer–Hilbert controversy.
(From https://en.wikipedia.org/wiki/Cantor%27s_paradise)
Best Answer
You are mixing several things from different parts of logic and set theory.
When we talk about decidable sets, the standard context is sets of natural numbers, or sets of tuples of natural numbers. And we are talking about them in the sense that there is a Turing machine that given a natural number will always stop and tell us whether or not this number is in the given set.
Turing machines can be thought of as computers. But not as physical computers. Those are ideal computers which are not limited by time, space, or energy constraints.
So given a natural number, $n$ is it even? Well, we need to go through all the natural numbers $k$ and ask if $k+k=n$. But in fact we only need to do it for $k\leq n$. So after at most $n$ steps we know if something is even or if it is not even. Therefore the even numbers form a decidable set.
Bit harder, but the set of prime numbers is also decidable. To judge whether or not $p$ is a prime we need to go through all the natural numbers $n<p$ and ask if there is some $k<p$ such that $n\cdot k=p$. Sure, it might take some time, but it's simple from a complexity point of view. After $p^2$ steps, we can decide if $p$ is a prime or not.
When we talk about uncountable sets, this is normally in the setting of some set theory (which may or may not be in the background of what you're doing). There we only talk about existence of bijection between your set and the fixed set $\Bbb N$. If there is such bijection, your set is countable, otherwise it is not.
You are not required to construct the bijection, or somehow provide a description for it. At least not in the classical settings of mathematics. You simply need to prove that there is one, or that there cannot be one. Cantor's diagonal argument shows that there cannot be a bijection between $\Bbb R$ and $\Bbb N$, so it is therefore the case that $\Bbb R$ is uncountable.
The two concepts are not really related. Decidable sets are countable (or finite), pretty much by the fact by definition as subsets of $\Bbb N$. But in the context of computability theory, the only sets of interest are subsets of $\Bbb N$ (or tuples of natural numbers, as we remarked above). So uncountability does not really come into play here.