(I)
Originally, there were symmetry groups: sets of symmetries of a mathematical object, like a figure in space or roots of a polynomial equation. A symmetry is a usually a transformation that preserves some chosen attributes of the mathematical object.
Then we distilled the abstract properties from symmetry groups to define an abstract group using axioms and without any reference to symmetry, only a binary operation and its properties. Thus we can study groups in their own right, and with the idea of isomorphism classes can compare the groups associated to different objects. (For example, a triangle has the same symmetry as a set of three elements, $S_3$, and a cube has the same rotational symmetry as a set of four elements, $S_4$.)
Going back to symmetry groups, one thing to notice about them is that there are often induced actions on various sets or structures associated to the original mathematical object. For example, given a symmetry group $G$ of a polyhedron in 3D space, there is an induced action of $G$ on the set $V$ of vertices, the set $E$ of edges, the set $F$ of faces, the set of space diagonals, the set of "flags," and so forth. This leads to the idea of group homomorphisms and group actions.
When every element of the group acts in a distinct way on a set $X$ (i.e. the map $G\to\mathrm{Perm}(X)$ is injective, in which case we call the action faithful), then $G$ is isomorphic to its image, so we can identify $G$ with a subgroup of $\mathrm{Perm}(X)$. Such a subgroup is called a permutation group. It is natural to ask, are "abstract groups" a more general class of groups than permutation groups? Could there abstract groups that are not isomorphic to permutation groups?
There aren't: every group acts faithfully on something. Let's say we have a group $G$ and want to describe an action of $G$ on some set $X$ but without assuming anything about $G$. First of all, we need a set to work with. We could pick a one-element set, but then $G$ acts trivially, which can't be a faithful action (unless $G$ is itself trivial of course). We could pick a two-element set, or some other set with a specific number of elements, but then there is no way to construct an action of $G$ on it without knowing anything about $G$ (indeed, depending on the sizes of $G$ and $X$, there may not be a nontrivial action at all!).
We have to create a set $X$ out of what we have available to us by hypothesis, but the only thing we have to work with is the group $G$ itself, so we simply use the regular action of $G$ on itself "by translation." That is, every $g\in G$ defines the permutation $\lambda_g:G\to G$ given by $\lambda_g(x)=gx$, and the map $g\mapsto\lambda_g$ is a homomorphism $G\to\mathrm{Perm}(G)$.
This is known as Cayley's Theorem. Often we interpret groups as symmetry groups, and symmetry groups can often look different (in particular, be "bigger") than the thing they are acting on, so describing the regular action as "translation" may in general be a bit unintuitive. However, the term "translation" makes perfect sense with the simplest examples: the real line $\mathbb{R}$, the circle $S^1$, or cyclic groups $C_n=\mathbb{Z}/n\mathbb{Z}$.
The real number line $\mathbb{R}$ may be identified with the group of translations of the number line. (In fact, this illustration works with $\mathbb{R}^2$, $\mathbb{R}^3$, etc.) And the circle $S^1$ may be identified with the group of 2D rotations. The cyclic group $C_n$ may be identified with the $n$th roots of unity, which form a $n$-gon in the circle group $S^1$ (itself viewed as a set of complex numbers), and this cyclic group is isomorphic to the group of 2D rotations by angles that are integer multiples of $2\pi /n$.
(II)
Yes, a symmetry group can itself be a geometrical object. Specifically, a Lie group is a smooth manifold. Indeed, $\mathbb{R}$ can be viewed as a symmetry group of itself (translations are the only transformations which preserve distances and do not reverse orientation); same with the circle group $S^1$ (rotations are the only transformations which preserve distances and orientation, again): both of these Lie groups are one-dimensional spaces.
Now let's go to two dimensions. Consider the group $\mathrm{Aff}(2,\mathbb{R})$ of orientation-preserving "affine transformations" of the plane $\mathbb{R}^2$. These are the transformations that preserve all distances (they are "isometries") and preserve orientation. Every such affine transformation is uniquely expressible as a translation and a rotation around the origin. There are a circle's worth of rotations around a point, which is $1$-dimensional, and there are an entire plane's with of translations, which is $2$-dimensional, for a total of $1+2=3$ dimensions! In fact, the space of all such affine maps of the plane is the same space as $S^1\times\mathbb{R}^2$! (The term is "diffeomorphic.")
(Note the boundaryless solid torus $S^1\times\mathbb{R}^2$ can itself be viewed as a Lie group, since $S^1$ and $\mathbb{R}$ are groups in their own right and it makes sense to take the direct product of groups. But while the groups $S^1\times\mathbb{R}^2$ and $\mathrm{Aff}(2,\mathbb{R})$ are the same space, they are not isomorphic as groups! One is abelian, the other isn't.)
Ultimately, "dimension" is a way to measure how many degrees of freedom there are, or for our purposes the number of real numbers needed in a parametrization. This is why it makes sense to add $1+2$ when determining the dimension of $\mathrm{Aff}(2,\mathbb{R})$ based on the decomposition into a knit product of $S^1$ and $\mathbb{R}^2$. Note these Lie groups are bona fide topological spaces with smooth structures, sometimes even metrics.
In particular, the 3D rotation group $\mathrm{SO}(3)$ may be viewed as a subset of the vector space of $3\times 3$ real matrices. Since this vector space has the a canonical norm (the Hilbert-Schmidt norm, associated to the Frobenius inner product in which the obvious canonical basis matrices are orthonormal), we can speak of the distance between points, and it turns out $\mathrm{SO}(3)$ is a $3$-dimensional submanifold of $M_3(\mathbb{R})$.
Here's a way to count the dimension. Every 3D rotation is a rotation around some axis by some angle. There are a 2D sphere's worth of oriented axes to choose from, and a circle's worth of angles to rotate by, for a total of $1+2=3$ dimensions. However, unlike with our previous example of $\mathrm{Aff}(2,\mathbb{R})$, it turns out $\mathrm{SO}(3)$ is not the same space as $S^2\times S^1$. Partly this is because a $0^{\circ}$ rotation around any axis is the same, and partly because rotation around an oriented axis by $\theta$ is the same as rotating around the opposite-pointing axis by the opposite angle $-\theta$.
However, it does turn out that $\mathrm{SO}(3)$ is "a bunch of circles arranged in the shape of a sphere" (the keyword here is "fiber bundle"), but they are in a sense twisted around the sphere. This is similar to how a Möbius band and a usual wristband can both be regarded as a bunch of line segments arranged in the shape of a circle, however the Möbius band has the line segment twist as it comes back around. But the twisting of circles around a sphere is harder to visualize.
(Also, it is probably easier to talk about the double cover $\mathrm{Spin}(3)$ of $\mathrm{SO}(3)$, since topologically it is just a $3$-sphere $S^3$ sitting inside four dimensions $\mathbb{R}^4$, and we have an associated "Hopf fibration" $S^1\to S^3\to S^2$.)
(In general, under nice conditions, there is a Lie group version of the "orbit-stabilizer theorem" in which if a group $G$ acts on a space $M$ with point stabilizer $S$, there is a fiber bundle $S\to G\to M$, which in particular implies $\dim G=\dim S+\dim M$. That applies here with $\mathrm{SO}(3)$ acting on $S^2$ with point stabilizer $\mathrm{SO}(2)\simeq S^1$.)
Note the symmetry group can be bigger-dimensional than the original space. The 3D rotation group $\mathrm{SO}(3)$ is $3$-dimensional and acts as the symmetry group of the $2$-dimensional sphere $S^2$ for example. Or if we consider the (orientation-preserving) affine transformations of 3D space instead of 2D, again every affine map is uniquely a translation and a rotation around the origin, for a total of $3+3=6$ dimensions, so in this sense the symmetry group of $\mathbb{R}^3$ is $6$-dimensional.
I said earlier that symmetry groups are comprised of symmetries, and symmetries are transformations that preserve a chosen set of attributes of a mathematical object. Well, we could ignore the metric / norm / inner product / topology etc. on $\mathbb{R}^3$ and only consider it as a vector space. In this case, the symmetries are all invertible linear maps, $\mathrm{GL}(3,\mathbb{R})$, again a subspace of the full vector space of matrices $M_3(\mathbb{R})$. In this case, $\mathrm{GL}(3,\mathbb{R})$ is in fact $9$-dimensional!
