I don't know if the following counts as "intuition", but perhaps it
can put things in some perspective.

It's a theorem that any rotation in $R^n$ can be written as the
composition of an even number of reflections through hyperplanes.
To construct a way of doing rotions by algebra (not matrix algebra,
but rather by multiplying vectors and similar geometric objects), it
is therefore enough to figure out two things:

(1) Given a vector $f$, how do we represent the reflection $R_f$
through the hyperplane orthogonal to $f$? We want $R_f(x)$ to be some
kind of product involving the vectors $x$ and $f$.

(2) The correspondence constructed in (1) should be such that for two
vectors $f$ and $g$, the algebraic object $fg$ (whatever that is)
should represent the composition $R_f \circ R_g$. Then any rotation
can be written as
$R_{f_1} \circ R_{f_2} \circ \cdots \circ R_{f_{2m}}$ and therefore represented by the object
$f_1 f_2 \cdots f_{2m}$.

For this to work, we have to give up the idea that the product of two
vectors must be a vector (as Hamilton famously struggled with for a
long time). Instead we let the vectors sit inside some bigger space of
"complex" (compound) objects that can be multiplied with each other
according to suitable rules (associative, but not commutative).

In the case of quaternions, these complex objects consist of a "scalar
part" plus a "vector part", and the construction (1) which makes
everything work is $R_f(x)=-fxf^{-1}$. To see that this is a
reflection, note that it's linear in $x$, it maps $f$ to $-f$, and it
maps vectors orthogonal to $f$ to themselves (since they satisfy
$fx=-xf$ because of the way quaternion multiplication is defined).

To do rotations in $R^n$ one constructs a so-called Clifford algebra,
consisting of complex objects with a "scalar part", a "vector part", a
"bivector part", a "trivector part", etc., up to an "n-vector part",
and the rules to multiply these objects are generated by the
requirement that for vectors $x$ and $y$, one wants $xy+yx=2 x\cdot y$
(or the negative of that, depending on convention). This makes
orthonal vectors anticommute, so that $R_f(x)=-fxf^{-1}$ gives a
hyperplane reflection, and everything works as desired.

I won't go into what a "k-vector" is, but the space of k-vectors
is $\binom{n}{k}$-dimensional, so that the whole Clifford algebra is
$2^n$-dimensional.
The Clifford product of two pure vectors is a scalar plus a bivector,
which is a little bit different from quaternions where vector times
vector equals a scalar plus a vector.
The full Clifford algebra for $R^3$ is 8-dimensional, and it's a bit
of a "coincidence" that one can actually manage just as well with the
4-dimensional quaternion algebra if one only wants to work with
rotations in $R^3$.