Let us remember the definition of faithful, transitive and free:
The action of a group $G$ on a set $X$ is called:
Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.
Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.
Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.
Now we can determine your three group actions:
The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.
The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.
The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.
Best Answer
What you wrote is correct, though a bit unclear. You could make things more clear as follows:
Let $x,y\in X$. There is a permutation $\sigma\in S_X$ mapping $x$ to $y$ (one could take the permutation $(x\,\,\,\, y)$). Since $\varphi$ is surjective, there is $g\in G$ such that $\varphi(g)=\sigma$. Then we have $g*x=\varphi(g)\cdot x=(x\,\,\,\,y)\cdot x=y$ as desired.
For the second question, indeed surjectivity is not needed. In fact, it is enough for the image of $\varphi$ to act transitively on $X$. For example, if $X$ has $n$ elements, then the cyclic subgroup of $S_n$ generated by the cyclic permutation $(1\,2\,\dots\,n)$ (or, in fact, by any $n$-cycle) acts transitively on $X$, and this subgroup has only $n$ elements. Another example is the subgroup $A_n$ of even permutations, which acts transitively on $X$.