The idea of an exponential is the continuous compounding of small actions. Suppose you start with an object $p$, perform an action on it $v$, and then add the result back to the original object. What happens if you instead take half as much action but do it twice? What about if you take one tenth the action but do it ten times? The exponential function tries to capture this idea:
$$\exp (\text{action}) = \lim_{n \rightarrow \infty} \left(\text{identity} + \frac{\text{action}}{n}\right)^n.$$
On a differentiable manifold there is no addition, but we can consider this action as pushing a point a short distance in the direction of the tangent vector,
$$``\left(\text{identity} + \frac{\text{v}}{n}\right)"p := \text{push }p\text{ by} \frac{1}{n} \text{ units of distance in the }v \text{ direction}.$$
Doing this over and over, we have
$``\left(\text{identity} + \frac{\text{v}}{n}\right)^n"p$ means push $p$ by $\frac{1}{n}$ units of distance in the $v$ direction, then push it again in the same direction you already pushed it, and keep doing so until you have pushed it $n$ times.
So long as $\frac{1}{n}$ is small enough that pushing points and vectors in a tangent direction makes sense, what we end out doing is pushing the point $p$ a total of $1$ unit of distance along the geodesic generated by $v$.
(Personally I use $\mathrm{SU}(2)$ strictly to denote the group of $2\times 2$ special unitary matrices and $\mathrm{Sp}(1)$ to denote the set of unit quaternions, but most people use them interchangeably since they are isomorphic Lie groups, like $\mathrm{U}(1)$ and $\mathrm{SO}(2)$.)
Do both the set of imaginary quaternions and the set of unit quaternions parameterize SU(2) via the exponential map?
This is kind of like asking "do both pairs of angles and the sphere parametrize the sphere via sines and cosines?" It's a weird thing to say. One uses sines and cosines to parametrize the sphere using two angles, sure, but one uses ... the identity function to parametrize a space with itself.
The unit quaternions are $\mathrm{Sp}(1)$, and each unit quaternion has a polar form given by the formula$\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}\,$ (just like complex numbers) where $\mathbf{u}$ is unit vector. Thus, the exponential function maps $\mathrm{Im}(\mathbb{H})=\mathbb{R}^3$ to the group $\mathrm{Sp}(1)$.
I don't understand the rest of what you're saying in question (1).
Is it true that each of the following three sets constitutes a Lie group: The set of unit quaternions, the set of imaginary quaternions, and the set of all nonzero quaternions?
The unit quaternions and nonzero quaternions are multiplicative groups and the imaginary quaternions are a vector space under addition, but the imaginary quaternions are certainly not closed under multiplication. Given two vectors $\mathbf{u}$ and $\mathbf{v}$, the product's real and imaginary parts (i.e. its scalar and vector components) are given by the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$, so the (quaternion) product of two vectors will itself be a vector if and only if they're orthogonal.
It's not clear to me how to define the appropriate metric for this to be the case.
There's already a metric on $\mathbb{H}$. It's given by
$$ |x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}|=\sqrt{x_0^2+x_1^2+x_2^2+x_3^2}. $$
So if you have a nice curve $\gamma:[0,1]\to\mathbb{H}$ its length can be calculated by $\int_0^1 |\gamma'(t)|\,\mathrm{d}t$, just as in any Euclidean space $\mathbb{R}^n$. Indeed, just as in $\mathbb{R}^2$ and $\mathbb{R}^3$, the length of an arc between two points on the unit sphere will be the angle between them.
For the unit quaternions, the tangent space at $1$ is just the imaginary quaternions; for non-unit quaternions, isn't the tangent space the same?
Nope. For inspiration, just look at $S^2$. Clearly antipodal points have the same tangent plane (interpreted as vector subspaces of $\mathbb{R}^3$ anyway), but otherwise the various tangent planes are not parallel and exist in different orientations.
Suppose $q:[0,1]\to\mathrm{Sp}(1)$ is a differentiable curve in the unit quaternions, and that $q(0)=1$. We have $q(t)\overline{q(t)}=1$; differentiating and then evaluating at $t=0$ yields $q'(0)+\overline{q'(0)}=0$, which is only possible if $q'(0)$ is purely imaginary since $x+\overline{x}=2\mathrm{Re}(x)$. But this argument relies on $q(0)=1$, i.e. on the fact we're look at the tangent space at $q=1$.
In general if $q(0)=q$ then instead you get $q'(0)\overline{q}+q\overline{q'(0)}=0$, or equivalently $2\mathrm{Re}(q'(0)\overline{q})=0$, so that $q'(0)\overline{q}$ is pure imaginary, or in other words $q'(0)$ is an an imaginary quaternion multiplied by $q$ from the right. Since $q^{-1}\mathbf{x}q$ is just $\mathbf{x}$ rotates, and $\mathbf{x}q=q(q^{-1}\mathbf{x}q)$, we could have equally well have said multiplying by $q$ from the left instead of the right.
In conclusion, the act of multiplying by $q$ slides (is a map from) the tangent space $T_1S^3$ to the tangent space $T_qS^3$. (Note $S^3$ is another term for $\mathrm{Sp}(1)$, just as $S^1$ is notation for $\mathrm{U}(1)$.)
If you want the exponential map $\exp:T_qS^3\to S^3$ in the Riemannian manifold sense, you can do a "transport of structure" argument: slide $T_qS^3$ to $T_1S^3$, apply the usual $\exp$ map, then slide back (so that $\exp$ of the zero vector in $T_qS^3$ yields $q$). Depending on if you slide using left or right multiplications, this yields $q\exp(\overline{q}x)$ or $\exp(x\overline{q})q$ (which are equal to each other).
Hopefully this answers (4) as well for you.
Best Answer
$T_{\log q} \mathbb R^3$ is the local tangent space defined at $\log q$, which is isomorphic to $\mathbb R^3$ defined by its tangent vectors $e_1$, $e_2$ and $e_3$.
In the picture you see that $d \exp_{\log q}$ is sending the tangent vectors $e_1$, $e_2$ and $e_3$ attached at $\log q$ to a tangent space at $q \in S^3$ (depicted as a plane) defined by tangent vectors $a_1$, $a_2$ and $a_3$ passing through $q$ where $a_i = d \exp_{\log q}( e_i) \in T_q S^3$.