For this answer I will identify $\mathbb{R}^3$ with the purely imaginary quaternions and although the quaternions are themselves a vector space, to avoid confusion I'll reserve the term 'vector' for members of $\mathbb{R}^3$.
There are 2 key identities which we will need, the first is the quaternion product of two vectors $a, b$
$$ab = -a \cdot b + a \times b$$
Which is a quaternion whose scalar part is the negative dot product of $a,b$ and whose vector part is the their cross product. This is derived by distributing terms, evaluating the products of imaginary units according to the quaternion multiplication table, and regrouping terms.
The second is a consequence of the first, for any unit vector $\hat n$ we have,
$$\hat n^2 = -\hat n \cdot \hat n + \hat n \times \hat n = -1 + 0 = -1$$
If you were to take a proof of Euler's formula for the complex numbers using Taylor series, replace every occurrence of $i$ with $\hat n$, and afterwards evaluate whether each step in that proof is still mathematically valid, you would find that they are, giving us Euler's formula for the quaternions,
$$e^{\theta \hat n} = cos(\theta) + sin(\theta) \hat n$$
One thing we can note is that any quaternion which lies in the $\{1,\hat n\}$-plane, the plane spanned by $1$ and $\hat n$, may be written in the form,
$$Ae^{\psi \hat n}$$
Where $A$ is a scalar magnitude and $\psi$ is a scalar angle, analogously to the polar form of complex numbers.
If we left-multiply such a quaternion by $e^{\theta \hat n}$,
$$e^{\theta \hat n} Ae^{\psi \hat n} = Ae^{(\psi + \theta) \hat n}$$
$A$ is a scalar and so commutes with $e^{\theta \hat n}$, and $\theta \hat n$ commutes with $\psi \hat n$ so we may combine the exponentials. We interpret the result as a rotation by an angle $\theta$ within the $\{1,\hat n\}$-plane.
If we right-multiply the same quaternion by $e^{\theta \hat n}$,
$$Ae^{\psi \hat n} e^{\theta \hat n} = Ae^{(\psi + \theta) \hat n}$$
We get the same result, a rotation by an angle $\theta$ within the $\{1,\hat n\}$-plane. $e^{\theta \hat n}$ commutes with quaternions in the $\{1,\hat n\}$-plane.
We can now try left-multiplying a unit vector $\hat p$ which is orthogonal to $\hat n$ by $e^{\theta \hat n}$,
$$e^{\theta \hat n} \hat p = \big( cos(\theta) + sin(\theta) \hat n \big) \hat p$$
$$= cos(\theta) \hat p + sin(\theta) \hat n \hat p$$
$$= cos(\theta) \hat p + sin(\theta) (-\hat n \cdot \hat p + \hat n \times \hat p)$$
$$= cos(\theta) \hat p + sin(\theta) \hat n \times \hat p$$
$\hat n \times \hat p$ is unit vector orthogonal to both $\hat n$ and $\hat p$, I'll define $\hat h = \hat n \times \hat p$. In search of a geometric interpretation we try a couple angles of $\theta$, when $\theta = 0$ the result is $\hat p$, when $\theta = \pi / 2$, the result is $\hat h$, it's then not too difficult to see that this is a rotation of $\hat p$ by an angle $\theta$ within the $\{\hat p,\hat h\}$-plane, which is also the plane orthogonal to $\hat n$ within 3D space. You can use the right hand rule to verify that this is a right-handed rotation around $\hat n$. We can note that any quaternion which lies within the $\{\hat p,\hat h\}$-plane may be expressed in the form,
$$Be^{\varphi \hat n} \hat p$$
Where $B$ is a scalar magnitude and $\varphi$ is a scalar angle, again analogously to the polar form of complex numbers.
If we instead right-multiply $\hat p$ by $e^{\theta \hat n}$,
$$\hat p e^{\theta \hat n} = \hat p \big( cos(\theta) + sin(\theta) \hat n \big)$$
$$= cos(\theta) \hat p + sin(\theta) \hat p \hat n$$
$$= cos(\theta) \hat p + sin(\theta) (-\hat p \cdot \hat n + \hat p \times \hat n)$$
$$= cos(\theta) \hat p - sin(\theta) \hat n \times \hat p$$
$$e^{-\theta \hat n} \hat p$$
We see that we instead get a rotation by $-\theta$ or a left-handed rotation around $\hat n$, $e^{\theta \hat n}$ does not commute with quaternions in the $\{\hat p,\hat h\}$-plane, it must be conjugated in order to reverse the order of multiplication with $\hat p$. We arrive at the following results for left and right multiplication of quaternions in the $\{\hat p,\hat h\}$-plane by $e^{\theta \hat n}$,
$$e^{\theta \hat n} Be^{\varphi \hat n} \hat p = Be^{(\varphi + \theta) \hat n} \hat p$$
$$Be^{\varphi \hat n} \hat p e^{\theta \hat n} = Be^{(\varphi - \theta) \hat n} \hat p$$
As $\{\hat n, \hat p, \hat h\}$ are mutually orthogonal unit vectors which span 3D space, it follows that $\{1, \hat n, \hat p, \hat h\}$ is an orthonormal basis for all quaternions. Any quaternion can be expressed as a linear combination of $\{1, \hat n\}$ summed with a linear combination of $\{\hat p, \hat h\}$, which we can write in the polar forms we found above,
$$q = Ae^{\psi \hat n} + Be^{\varphi \hat n} \hat p$$
For fixed $\hat n$ and $\hat p$, the 4 degrees of freedom in a general quaternion are contained in $A, \psi, B, \varphi$, a set of polar coordinates for each plane. Expressed in this form it is easy to see the effects of left and right multiplication of a general quaternion by $e^{\theta \hat n}$,
$$e^{\theta \hat n} q = Ae^{(\psi + \theta) \hat n} + Be^{(\varphi + \theta) \hat n} \hat p$$
$$q e^{\theta \hat n} = Ae^{(\psi + \theta) \hat n} + Be^{(\varphi - \theta) \hat n} \hat p$$
Left-multiplication rotates both components of $q$ by an angle $\theta$ within their respective planes, while right-multiplication rotates the component of $q$ within the $\{1, \hat n\}$-plane by an angle $\theta$ within that plane and the component of $q$ in the $\{\hat p, \hat h\}$-plane by an angle $-\theta$ within that plane.
If we wanted to rotate only one component of $q$ instead of both of them, we can use the two sided multiplications with half-angles, these result in the rotations within one plane adding up to a total of $\theta$ and the rotations within the other plane cancelling out to $0$,
$$e^{\theta \hat n / 2} q e^{\theta \hat n / 2} = e^{\theta \hat n / 2} \big( Ae^{\psi \hat n} + Be^{\varphi \hat n} \hat p \big) e^{\theta \hat n / 2}$$
$$= Ae^{\theta \hat n / 2} e^{\psi \hat n} e^{\theta \hat n / 2} + Be^{\theta \hat n / 2} e^{\varphi \hat n} \hat p e^{\theta \hat n / 2}$$
$$= Ae^{\theta \hat n / 2} e^{\psi \hat n} e^{\theta \hat n / 2} + Be^{\theta \hat n / 2} e^{\varphi \hat n} \hat e^{-\theta \hat n / 2} \hat p$$
$$= Ae^{(\psi + \theta) \hat n} + Be^{\varphi \hat n} \hat p$$
$$e^{\theta \hat n / 2} q e^{-\theta \hat n / 2} = e^{\theta \hat n / 2} \big( Ae^{\psi \hat n} + Be^{\varphi \hat n} \hat p \big) e^{-\theta \hat n / 2}$$
$$= Ae^{\theta \hat n / 2} e^{\psi \hat n} e^{-\theta \hat n / 2} + Be^{\theta \hat n / 2} e^{\varphi \hat n} \hat p e^{-\theta \hat n / 2}$$
$$= Ae^{\theta \hat n / 2} e^{\psi \hat n} e^{-\theta \hat n / 2} + Be^{\theta \hat n / 2} e^{\varphi \hat n} \hat e^{\theta \hat n / 2} \hat p$$
$$= Ae^{\psi \hat n} + Be^{(\varphi + \theta) \hat n} \hat p$$
We can see that by using the two-sided multiplications with half-angles we can successfully isolate the rotation in the $\{1, \hat n\}$-plane if we don't conjugate second factor of $e^{\theta \hat n}$, and the rotation in the $\{\hat p, \hat h\}$-plane if we do conjugate it.
Now if our goal is to rotate a vector $v$ within 3D space by an angle $\theta$ around axis $\hat n$, then we would like the component of $v$ in the $\{\hat p, \hat h\}$-plane to rotate by an angle $\theta$, and the component parallel to $\hat n$ i.e. the component in the $\{1, \hat n\}$-plane to remain fixed, we certainly don't want it to rotate into some non-physical 4th dimension, so we cannot use either of the one-sided multiplications as they do in general rotate the component within the $\{1, \hat n\}$-plane. We must use the conjugated two-sided multiplication to achieve this as we found above, giving us our final vector rotation formula,
$$v' = e^{\theta \hat n / 2} v e^{-\theta \hat n / 2}$$
Expanding out the exponentials gives us the familiar format of the rotation formula,
$$v' = \big(cos(\theta / 2) + sin(\theta / 2) \hat n \big) v \big(cos(\theta / 2) - sin(\theta / 2) \hat n \big)$$
So to answer your questions, you don't need to use the cos-sin representation of the rotation quaternions, they have exponential representations as well. You were on the right track by looking for the connection to 2D rotation using complex exponentials, we can view quaternion multiplication as two rotations in two orthogonal planes which are individually not too dissimilar from the 2D case. If you want to rotate a vector around an axis, then you want don't want the component of that vector which is parallel to the rotation axis to change, but the one-sided transformations will rotate this component into the real axis in general, so we need to strategically use two of them in order to get this unwanted rotation to cancel out, and that's why we need the two-sided transformation.
