Is generated group by $\{a,b,c,c^2\}$ same as group generated by $\{a,b,c\}$? I think the answer is YES. But here is a paragraph of J. Wolf's Book:
Let $\triangle_8$ denote the regular octahedron (6 vertices, 12 edges, 8 faces; the faces are equilateral triangles). Let $X$ be rotation of $\pi/2$ about the line through $v_1$ and $v_6$, $ P = X^2$. Let $Q$ be rotation of $3\pi/2$ about the line through $v_2$ and $v_4$. Then $PQ = QP$ is rotation of $\pi$ about the line through $v_3$ and $v_5$. ... One checks that $\{P, Q, A, X\}$ generates the group of all symmetries of $\triangle_8$, and that $X$ can be changed to another generator $R$.
What is the problem?
The groups generated by $\{a,b,c,c^{2}\}$ and by $\{a,b,c\}$ are indeed the same, as you noticed, since $c^{2} \in \langle a,b,c\rangle$.
Wolf is going through these steps to get a nice presentation of $\Delta_{8}$, in particular he wants a presentation in terms of generators all having order 2. You have that $\Delta_{8}$ is generated by $\{P,Q,A,X\}$; this is easy to verify. While it is true that $\Delta_{8}$ is generated by $\{Q,A,X\}$, this is not the presentation that Wolf wants (I'm guessing because it lacks symmetry in terms of the roles the generators play, and also he wants to emphasize $\langle P,Q,A\rangle$ as a subgroup).
Now that you have verified that $\{P,Q,A,X\}$ generate $\Delta_{8}$, Wolf wants to replace $X$ with $R$; the idea is you do this so that $X \in \{P,Q,A,R\}$, so the whole group is still generated, and so that $R$ has the property that $R^{2} = 1$ (though Wolf does not say how $R$ is defined). I'm guessing that this makes is so the group is now NOT generate by any proper subset of $\{P,Q,A,R\}$.
This is now a "nice" presentation for $\Delta_{8}$ since certain important subgroups can be expressed easily, and all generators have the same order.
Nowhere is Wolf trying to obtain the most efficient generating set for this group (in fact, this group is isomorphic to $S_{4}$ and can therefore be generated by two elements).