I've just read the first few pages of Combinatorial Group Theory by Magnus, Karrass, and Solitar, and based on their definitions there, and more specifically, the reasoning given in the hint to exercise $5$(c) (on p. 9), I don't see why $\langle a ; a^2 \rangle$ (or even $\langle a ; a \rangle$, for that matter) shouldn't be regarded as a presentation of $C_4$.
Exercise $4$ (also on p. 9), asks the reader to establish that "if the word $K$ is derivable from the relators $M, N, \dots,$ and $M, N, \dots$ are derivable from the relators $P, Q, R, \dots,$ then $K$ is derivable from $P, Q, R, \dots\,$".
Exercise $5$(b) asks the reader to establish that $\langle a; a^4 \rangle$ is a presentation of $C_4$. The sequence below
$$a^4,a^4a^{-2},a^2,a^2a^{-2},1$$
...shows that the word $a^4$ is derivable (using the operations stipulated on p. 6) from the relator $a^2$ (and is hence itself a relator).
Of course, I realize that $\langle a; a^2 \rangle$ is also a presentation of $C_2$, and that $C_2 \neq C_4$.
And pretty much the same reasoning would lead one to conclude that $\langle a; a^2 \rangle$ is a presentation of $C_{2n}$, and $\langle a; a \rangle$ is a presentation of $C_n$, for any $n \in \{2, 3, \dots\}$. If so, a presentation of a group $G$ could not be considered (necessarily) a description, or specification, of $G$; so what's its purpose?
Therefore, I conclude that something must be missing, either from the authors' definition of a group presentation (p. 7)1, or from my grasp of it.
EDIT: To clarify, let me say more about the origin of my question, namely the hint to exercise $5$(c) (already alluded to at the beginning of this post).
Exercise $5$(c) asks
Give a presentation for [the cyclic group of order 4] using one generating symbol $b$ and two defining relators, neither of which is $b^4$ or $b^{-4}$.
The hint for this question says:
...use $b^8$ and $b^{12}$ as defining relators, and show that $b^4$ can be derived. Hence, by [exercise 4, cited above], $b^8$ and $b^{12}$ are a set of defining relators.
I have no problem following this hint, but I can just as well carry out exactly the same steps not just for $(b^8, b^{12})$, but for any pair $(b^n, b^{n+4})$ with $n \in \mathbb{Z} \backslash \{-1, 0, 1\}$, like so: $$b^4,b^4b^n,b^{n+4},b^{n+4}b^{-(n+4)},1$$
And if this maneuver somehow renders $\langle b; b^n, b^{n+4}\rangle$ (for any $n \in \mathbb{Z} \backslash \{-1, 0, 1\}$) a presentation for $G$, then a similar maneuver would render $\langle a; a^2\rangle$, and even $\langle a; a\rangle$ a presentation for $G$.
I realize that this conclusion is nonsense. I'm trying pinpoint the misstep in the reasoning.
1 For completeness, this definition goes like this: "A word $R(a, b, c)$ which defines the identity element $1$ in [a group] $G$ is called a relator. … If every relator [of a group $G$] is derivable from the relators $P, Q, R, \dots,$ then we call $P, Q, R, \dots$ a set of defining relators or a complete set of relators for the group $G$ on the generators $a, b, c, \dots .$ If $P, Q, R,\dots$ is a set of definitng relators fro the group G on the generators $a, b, c,\dots$ we call $$\langle a, b, c, \dots; P(a, b, c, \dots), Q(a, b, c, \dots), R(a, b, c, \dots), \dots\rangle$$ a presentation of $G$ and write $$G = \langle a, b, c, \dots; P, Q, R, \dots \rangle\,.$$
When you write $G=\langle a;a^2\rangle$, this is meant that $G$ is the group generated by an element $a$ such that $a^2=1$. Notice that this means that $a=a^{-1}$. So $G$ is just $\{1,a\}$. For any $n\in\mathbb{N}$, the group $G=\langle a;a^n\rangle$ is just the group, $\{1,a,a^2,\ldots,a^{n-1}\}$, which is $C_n$. In your case you have if $a^2=1$, then $(a^2)^2=a^4=1$, but not the other way around. This is what @studiosus was getting at (I don't think his/her intent was to insult).