The well-ordering principle has always been considered to be highly unconstructive, as far as I know. However, I think intuitionistic mathematics can be compatible with the existence of a well-ordering of $\mathcal N = \{ \alpha : \mathbb N \to \mathbb N \}$ (or other uncountable sets like $\mathbb R$), which Brouwer and others did not seem to accept.
I know that a full, positive version of well-ordering would imply some omniscience principle, so let's begin with a weaker property for orderings $<$ on a set $X$ that are total in the sense $$\lnot (\lnot x = y \land \lnot x \lt y \land \lnot y \lt x).$$ Let's call it a well-ordering if
$$ WO(<) \equiv \lnot (\exists f : \mathbb N \to X) (\forall n) (f(n+1) < f(n)).$$ (This is probably not the most efficient principle, just note that it can't be the usual positive one)
Question: is the principle that $\mathcal N$ can be well-ordered consistent with the principles of intuitionistic analysis? (continuity principle and bar theorem or Church thesis)
I'll now show my argument that says that Brouwer could have accepted this principle, unless of course there's a serious issue.
I begin with any element of $\mathcal N$. Every time I construct a proof that an element $\alpha \in \mathcal N$ is distinct from all the previous elements, I add it to the end (i.e., postulate $\beta < \alpha$ for every previous $\beta$).
I might instead come up with a whole sequence $\{\alpha_n\} \subseteq \mathcal N$ and a proof that no $\alpha_n$ has appeared previously. Then I add the whole sequence at the end (i.e., I postulate $\beta < \alpha_0 < \alpha_1 < \dots < \alpha_n < \dots$ for all previous $\beta$).
Now, it might appear that diagonalization would lead to problems. But diagonalization is just a particular way of constructing a new $\alpha$ that has not previously appeared.
A similar objection is that the resulting order type would be $\omega^2$, and thus the order isomorphism $\omega^2 \to \mathcal N$ would allow further diagonalization, contradicting totality. However, this surveyable ordinal has certain positive properties (like trichotomy), that prevent the construction isomorphism.
Note that every new element that I construct will immediately be compared with all the previously constructed elements. Thus the resulting ordering possibly has the desired properties (including $WO$).
A simpler approach is to consider an uncountable subset of $\mathbb N$ with the usual order relation, I think it has the properties $WO$ and trichotomy, but such a subset must be undecidable. The details likely will depend on the specific subset.
Summing up: I don't understand why the well-ordering principle for uncountable sets is necessarily not constructive. But I haven't been able to find a source that claims otherwise.
Perhaps the negative versions of well-ordering are of very limited use, and that's why they are not discussed? Or are they inconsistent with certain constructive principles?
Yes, this principle is indeed consistent with bar induction and the continuity principle. It is separately consistent with Church’s Thesis.
Consider the following:
This principle is indeed consistent with bar induction and continuity. It is also obviously consistent with Church’s Thesis. Finally, it implies that $\mathcal{N}$ can be “well-ordered” in your sense.
Assume Weak Church’s Thesis.
We say a natural number $n$ represents a function $f : \mathbb{N} \to \mathbb{N}$ if $n$ is the index of the computable function $f$ (using a canonical enumeration of the partial computable functions, eg with Kleene’s T predicate). Write this as $R(n, f)$. The above principle states $\forall f \neg \neg \exists n (R(n, f))$.
Now recall the following constructive theorem:
The proof of the above is by induction.
We say that $n$ is the canonical representative of $f$ if $n$ is the smallest $n$ such that $R(n, f)$. Write this as $CR(n, f)$. Then we have $\forall f \neg \neg \exists n (CR(n, f))$.
We are now ready to define our “well-order” on $\mathcal{N}$. We say $f < g$ if and only if $\exists n \exists m (n < m \land CR(n, f) \land CR(m, g))$.
It’s straightforward to show that $<$ is irreflexive and transitive. It’s also straightforward to show it is a “well-ordering”. For if we have a function $h : \mathbb{N} \to \mathcal{N}$ such that $\forall n (h(n + 1) < h(n))$, then each $h(n)$ has a canonical representative. Define $g(n)$ to be the canonical representative of $h(n)$. Then $g : \mathbb{N} \to \mathbb{N}$ is strictly decreasing; contradiction.