The way I understand it is as follows. The most general formula

$
(\neg A \; \to \; B \vee C) \;\; \to \;\;
((\neg A \; \to \; B) \vee (\neg A \; \to \; C))
$

is not derivable in IPC, because it is not an intuitionistic truth, like you said.

In fact, we can find `almost intuitionistic' Brouwerian counterexamples outside of IPC. For instance if we look at HA + MP and take

$R(n):= 1 \ \mbox{if the first block of 99 consecutive 9's}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{in the decimal expansion of $\pi$ ends at place $n$}$

$R(n):= 0\ \ \mbox{else}$

$A=\forall n [R(n)=0]$ $\ \ \ B= \exists n [R(2n)=1]$
$\ \ \ C= \exists n [R(2n+1)=1]$

Then we see that $\neg A\to B\vee C$ (using MP). But both $(\neg A \; \to \; B)$ and $(\neg A \; \to \; C)$ are elusive. The interesting thing here is, that if HA really proves $\neg A$, then without MP already HA proves `there is $m$ with $R(m)=1$'. This since HA is closed under Markov's Rule, therefore proving $\neg\neg\exists n[P(n)]$ for decidable $P$ in HA allows one to deduce something about the structure of the proof, from which one can find a proof in HA of $\exists n [P(n)]$. This is a proof-theoretical property of the formal system HA.

Similarly, if in IPC we have formulas $A,B,C$ such that IPC proves $(\neg A \; \to \; B \vee C)$, then we know from the proof-theoretical properties of IPC that we can either find a proof in IPC of $(\neg A \; \to \; B)$ or of $(\neg A \; \to \; C)$.

So both computationally and semantically, this Harrop's Rule (also called Independence of Premise Rule, or IPR) being admissible for IPC should be seen as a statement about certain properties of the formal system IPC.

A nice reference is Rosalie Iemhoff's paper *On the rules of intermediate logics*.

**Addendum**:
It just occurred to me that the question might be asking for an informal semantical/computational explanation why IPC has this property. In that case, consider that in general we cannot extract numerical existence from a negated formula $\neg A$. Therefore, if for a formula $A$ in IPC we deduce that $\neg A$ implies $\exists n\in\{0,1\} [(n=0\to B)\wedge (n=1\to C)]$, then we must have a means to pinpoint this $n$ *without* having to first prove $\neg A$. And that is precisely Harrop's rule IPR.

This is also why we need Markov's Principle (MP) in the counterexample above: to extract numerical existence from the negated formula $\neg\forall n[R(n)=0]$ which is equivalent to $\neg\neg\exists m [R(m)=1]$