Here's one way. The group of unit quaternions is isomorphic to the special unitary group $\text{SU}(2)$, the group of $2 \times 2$ unitary complex matrices with determinant $1$. This group acts on $\mathbb{C}^2$ in the obvious way, and so it also acts on lines in $\mathbb{C}^2$. (These are complex lines, so they have real dimension $2$.) The space of lines in $\mathbb{C}^2$ is the complex projective line $\mathbb{CP}^1$, and it turns out there is a natural way to think about this space as a sphere - namely, the Riemann sphere. There is a beautiful projection which is pictured at the Wikipedia article which shows this; essentially one thinks of $\mathbb{CP}^1$ as $\mathbb{C}$ plus a "point at infinity" and then projects the latter onto the former in a way which misses one point.
So $\text{SU}(2)$ naturally acts on a sphere, and as it turns out it naturally acts by rotations. This describes the famous 2-to-1 map $\text{SU}(2) \to \text{SO}(3)$ which allows quaternions to describe 3D rotations.
The geometric interpretation of quaternion multiplication is fundamentally 4-dimensional (unlike quaternion conjugation, which can be considered as an action on $\Bbb{R}^3$).
Let's start with an easy case. Say $q=a+bi$ with $b \neq 0$, $a^2+b^2=1$. That is, $q$ is a non-real unit quaternion in the subalgebra of $\Bbb{H}$ generated by $i$. What affect does multiplying by $q$ have on an arbitrary quaternion $r$?
First of all, if $r$ also lies in the subalgebra generated by $i$, then we can consider multiplication of $q$ and $r$ to be ordinary complex multiplication; that is, multiplication by $q$ rotates $r$ by $\theta$ in the $\{1, i\}$ plane.
Secondly, if $r$ lies in the orthogonal complement of that subalgebra, $r=cj+dk$, we can write $r=(c+di)j=j(c-di)$. The first of these representations can be used to left-multiply by $q$ via ordinary complex multiplication; the second one can be used to right-multiply. In either case, multiplying by $q$ rotates $r$ by $\theta$ in the $\{j, k\}$-plane; however, the sign difference means that the two multiplications rotate in opposite directions from each other.
We can then find the effect of multiplying $q$ by an arbitrary quaternion by projecting that quaternion into these two planes. That is, an arbitrary quaternion will have its $\{1, i\}$-projection and $\{j, k\}$-projection both rotated by $\theta$ when it is multiplied by $q$; the direction of the $\{j, k\}$-rotation, but not of the $\{1, i\}$-rotation, will be affected by whether we're left-multiplying or right-multiplying by $q$.
The general case works similarly. For any unreal unit quaternion $q$ that makes an angle $\theta$ with the real axis, multiplication by $q$ rotates by $\theta$ in the $\{1, q\}$-plane, and also rotates by $\theta$ in its orthogonal complement. The direction of the first rotation is fixed, but the direction of the second rotation depends on whether we're multiplying by $q$ on the left or on the right. You can see this just by noticing that any unreal quaternion generates a 2-dimensional subalgebra isomorphic to $\Bbb{C}$, making the previous few paragraphs work in general after some relabeling.
This also gives you a way to see why quaternion conjugation works the way it does on $\Bbb{R}^3$. If $q$ makes an angle $\theta$ with the real axis, then the map $r \mapsto qrq^{-1}$:
- is the identity map in the $\{1, q\}$-plane, since that plane forms a commutative subalgebra of $\Bbb{H}$
- involves a rotation by $2\theta$ in its orthogonal complement. Both $q$ and $q^{-1}$ rotate quaternions in $\{1, q\}^{\perp}$ by $\theta$. If they were multiplied on the same side, those rotations would have to cancel out; since left-multiplication behaves oppositely to right-multiplication, this means they must reinforce each other. Since the orthogonal complement of $\{1, q\}$ is orthogonal to $1$, it is pure imaginary, so we've reproduced the fact that quaternion conjugation corresponds to a double-angle rotation in $\Bbb{R}^3$ (when identified with $\Im(\Bbb{H})$).
Note also that multiplying by a general quaternion involves scaling by the norm of that quaternion, but conjugation conveniently causes the norms of $q$ and $q^{-1}$ to cancel.
Best Answer
Don't try interpret quaternion multiplication $q_1 q_2$ as immediatedly connected to rotations; it's more indirect.
The operation which performs a rotation is $q u q^*$, where $u$ is a pure vector quaternion (to be rotated), and $q$ is a unit quaternion $$ q = \cos(\alpha/2) + \sin(\alpha/2) \, n , $$ where $n$ is a unit vector giving the axis of rotation, and $\alpha$ is the angle of rotation. (And $q^* = \cos(\alpha/2) - \sin(\alpha/2) \, n$ is the conjugate.)
Then a general quaternion (a constant times a unit quaternion) can be thought of as representing a linear transformation which is a combination of a rotation and a scaling, and the multiplication of two quaternions is an algebraic way of computing the composition of two such linear transformations: $$ q_1 \bigl( q_2 u q_2^* \bigr) q_1^* = (q_1 q_2) u (q_1 q_2)^* . $$ For example, a pure vector (unit) quaternion performs a 180 degree rotation around the corresponding axis, but when you compose two 180 degree rotations you don't get a new 180 degree rotation, unless the two axes of rotation happen to be orthogonal. So it makes perfect sense to have a scalar part (nonzero in general) in the quaternion product; it is related to the angle of the composed rotation.