I think a lot of the confusion around your question comes from the fact that you don't really have a 'chain' of the sort you're talking about, because different senses of the word 'transform' are being used in each case.
Very broadly, there's little distinction between 'points' and 'vectors' in the sense that you're specifying; each point can be identified with the vector from the origin to that point. Viewed from this perspective, the transformation process you're talking about that takes the form of a map (or, if you prefer, a function) from $\mathbb{R}^n\rightarrow\mathbb{R}^n$ (here $\mathbb{R}^n$ just means the space of $n$-dimensional vectors, and $\mathbb{R}$ means we're talking about vectors over the real numbers). Each vector $\vec{v}$ gives rise to the transformation $T_{\vec{v}}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ given by $T_{\vec{v}}(\vec{w}) = \vec{v}+\vec{w}$. Note that 'applying' two different vectors $\vec{u}$ and $\vec{v}$ to the same point $\vec{w}$ is the same as applying their sum: $T_\vec{u}(T_\vec{v}(\vec{w})) = T_\vec{u}(\vec{v}+\vec{w}) = \vec{u}+(\vec{v}+\vec{w}) = (\vec{u}+\vec{v})+\vec{w} = T_{\vec{u}+\vec{v}}(w)$.
Now, it's worth noting here that I used '$n$' above — vectors aren't limited to just 3-dimensional space. The example you gave, for instance, is a two-dimensional vector, and vectors make sense in five dimensions, ten dimensions, and even infinitely many dimensions (some caveats apply!). On the other hand, quaternions only${}^* $ work in 3 dimensions (${}^* $: and 4, sort of, but that's a more complicated story) and they don't produce all transformations of vectors in 3 dimensions; they only represent rotations of 3-dimensional space. Each quaternion $\mathbf{q}$ gives rise to a transformation $T_{\mathbf{q}}: \mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $T_{\mathbf{q}}(\vec{v}) = \mathbf{q}\vec{v}\mathbf{q}^{-1}$ , where all of the multiplications are quaternion multiplications and we're representing the vector as a 'purely imaginary' quaternion (one with zero scalar part); even though you're multiplying quaternions here, the result will still be a vector (that is, the scalar part will be zero). And much like with the vector 'transformation' above, composing transformations — that is, applying multiple quaternions to a vector, one after the other — corresponds to applying their product: $T_{\mathbf{p}}(T_{\mathbf{q}}(\vec{v})) = T_{\mathbf{p}}(\mathbf{q}\vec{v}\mathbf{q}^{-1}) = \mathbf{p}(\mathbf{q}\vec{v}\mathbf{q}^{-1})\mathbf{p}^{-1} = (\mathbf{p}\mathbf{q})\vec{v}(\mathbf{q}^{-1}\mathbf{p}^{-1}) = (\mathbf{p}\mathbf{q})\vec{v}(\mathbf{p}\mathbf{q})^{-1} = T_{\mathbf{p}\mathbf{q}}(\vec{v})$
Now, you may have noticed that even though these two examples are different, they do share a bit of flavor: in both cases, we can say that an object of the given type (a vector or a quaternion) gives rise to a transformation (or a map) from a space to itself, such that composing the transforms corresponds to adding (in the first case) or multiplying (in the second) the objects. This concept comes up often enough that mathematicians have come up with a special name for it, the group action; you can find more about it under that name. A group basically just means 'a bunch of abstract things with some special operator for turning two things in the group into a third, satisfying certain conditions'; here, our groups are the group of all vectors (that is, the things in the group are vectors and the operator is vector addition) and the group of all quaternions (where the things in the group are quaternions and the operator is vector multiplication), and the spaces they're acting on are either the space $\mathbb{R}^n$ of $n$-dimensional vectors in the first case or the specific space $\mathbb{R}^3$ of $3$-dimensional vectors in the second. What you're asking is basically whether there's some other group that acts on the space of all quaternions $\mathbb{H}$ in this same way. (Note that we call the quaternions '$\mathbb{H}$' in honor of their discoverer Hamilton).
Unfortunately, it turns out that there's an almost-trivial example, and it's closely related to your first example: any group always acts on itself, with the 'transformation' given by the group operation itself. In the case of vectors, this is exactly your first example; every vector $\vec{v}$ gives rise to a transformation from $\mathbb{R}^n\rightarrow\mathbb{R}^n$ given by the 'group operation' vector addition. In the case of the quaternions, it means that the quaternions transform themselves; for each quaternion $\mathbf{p}$ there's a transformation $T_{\mathbf{p}}:\mathbb{H}\rightarrow\mathbb{H}$ given by $T_{\mathbf{p}}(\mathbf{q}) = \mathbf{p}\mathbf{q}$. If you're wondering about other groups with an action on the quaternions - that, unfortunately, I'm not sure of. Rotations of the quaternions (seen as a general 4-dimensional space) can actually be specified by a pair of unit quaternions $\mathbf{p}$, $\mathbf{q}$ where their action on a given point in that space (that is, a quaternion) $\mathbf{h}$ is given by $T_{\mathbf{p},\mathbf{q}}(\mathbf{h}) = \mathbf{p}\mathbf{h}\mathbf{q}$; Wikipedia's page on 4-dimensional rotations has some more details on this, but I'll warn you in advance that they start to get very boggy very quickly.
Hopefully this gives you, if not quite an answer to your question, some more information on the sort of thing you're looking for! These concepts (representations and actions) are very core in a lot of areas in mathematics and I highly encourage you to dig more into them.