How can some statements be consistent with intuitionistic logic but not classical logic, when intuitionistic logic proves not not LEM? I've heard that some axioms, such as "all functions are continuous" or "all functions are computable", are compatible with intuitionistic type theories but not their classical equivalents. But if they aren't compatible with LEM, shouldn't that mean they prove not LEM? But not LEM means not (A or not A) which in particular implies not A - but that implies (A or not A). What's gone wrong here?

If $A$ is a sentence (ie has no free variables), then your reasoning is correct and in fact $\neg (A \vee \neg A)$ is not consistent with intuitionistic logic.
However, all the instances of excluded middle that are contradicted by the statements "all functions are continuous" and "all functions are computable" are for formulas of the form $A(x)$ where $x$ is a free variable. To give an explicit example, working over Heyting arithmetic (HA), let $A(n)$ be the statement that the $n$th Turing machine halts on input $n$. Then, it is consistent with HA that $\forall n\;A(n) \vee \neg A(n)$ is false. That is, $\neg (\forall n \; A(n) \vee \neg A(n))$ is consistent with HA, and is in fact implied by $\mathsf{CT}_0$ (essentially the statement that all functions are computable). Note that even in classical logic this doesn't directly imply $\neg A(n)$, which would be equivalent to $\forall n \; \neg A(n)$. What we could do in classical logic is deduce $\exists n \; \neg (A(n) \vee \neg A(n))$ and continue as before, but this does not work in intuitionistic logic.
