Let $g$ be a generator of your group. Then every element is of the form $g^i$ for some integer $i$ satisfying $0 \leq i < 4k$. If $g^i$ has order $4$, then $(g^i)^4 = g^{4i} = g^0$, so $4k$ divides $4i$, i.e., $k$ divides $i$. Thus your element is in the subgroup generated by $g^k$; this is a cyclic subgroup of order $4$. The question thus reduces to finding how many elements have order $4$ in the cyclic subgroup of order $4$; this should be easier.
I have started from the stage where I got stuck in proving the above lemma. It is easy to show what I just mentioned in the edit is that $\text {Ord}\ (ab)\ \big |\ \text {lcm}\ \left (\text {Ord}\ (a), \text {Ord}\ (b) \right ).$ To prove equality we need to prove the other way round which is not true for arbitrary finite groups even if $a$ and $b$ commute. We are so lucky that the other part is true for our case. Why? Lets discuss.
Before proving the required result I noticed that if we can prove the following lemma we are through.
Lemma $:$ Let $\sigma, \tau \in S_n$ be two disjoint cycles. Then $\text {Ord}\ (\sigma \tau ) = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
For proving the equality in the lemma let us first introduce the following definition.
Let $\rho = (a_1,a_2, \cdots , a_r) \in S_n$ be an $r$-cycle. Then the support of $\rho$ is denoted by $\text {Supp}\ (\rho)$ and it is defined as $\text {Supp}\ (\rho) = \{a_1,a_2, \cdots , a_r \}.$ So $\text {Supp}\ (\rho)$ consists of those points in $\{1,2, \cdots, n \}$ which are disturbed by the operation of $\rho.$
Observation $:$ If $\rho,\rho' \in S_n$ are two cycles inverses of each other then $\text {Supp}\ (\rho) = \text {Supp}\ (\rho').$ (Because inverse cycles fix same points).
Now let us take two disjoint cycles $\sigma , \tau \in S_n.$ On contrary let us assume that $\text {Ord}\ (\sigma \tau) = m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Then it is easy to see that $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Let us assume that $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ for otherwise $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ),$ a contradiction to our assumption. Since fixed points of $\sigma$ and $\tau$ are respectively fixed points of $\sigma^m$ and $\tau^m$ respectively it follows that $\text {Supp}\ (\sigma^m) \subseteq \text {Supp}\ (\sigma)$ and $\text {Supp}\ (\tau^m) \subseteq \text {Supp}\ (\tau).$ Since $\sigma$ and $\tau$ are disjoint cycles so we have $\text {Supp}\ (\sigma) \cap \text {Supp}\ (\tau) = \varnothing.$ Hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) = \varnothing.\ \ \ \ (*)$
Now since $\text {Ord}\ (\sigma \tau) = m$ so we have $$\begin{align*} (\sigma \tau)^m & = \text {id} \implies \sigma^m \tau^m = \text {id} \implies \sigma^m = (\tau^m)^{-1} \end{align*}$$
So $\sigma^m$ is the inverse of $\tau^m.$ So from our Observation it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m).$ Since $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m) \neq \varnothing$ and hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) \neq \varnothing,$ which contradicts $(*).$ That implies either $\sigma^m = \text {id}$ or $\tau^m = \text {id}.$ But if one of $\sigma^m$ or $\tau^m$ is identity then by using the equation $\sigma^m \tau^m = \text {id}$ we find that the other is also an identity. So we must have $\sigma^m = \tau^m = \text {id}.$ This implies $\text {Ord}\ (\sigma)\ \big |\ m$ and $\text {Ord}\ (\tau)\ \big |\ m.$ But it means that $\text {lcm}\ \left ( \text {Ord}\ (\sigma),\text {Ord}\ (\tau) \right )\ \bigg |\ m,$ which is a contradiction to our assumption that $m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence our assumption is false. So $m \geq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ But since $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right )$ it follows that $m \leq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence combining these two inequalities it follows that $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
QED
Best Answer
For $C_6$, observe that $\gamma_C(1)=1$, $\gamma_C(2)=1$, $\gamma_C(3)=2$ and $\gamma_C(6)=2$. The order of an element $(a,b)$ with $a,b\in C_6$ is the lcm of their respective orders and hence is again a divisor of $6$. Therefore we have $\gamma_G(1)=\gamma_C(1)^2=1$ as $(a,b)$ has order $1$ iff both $a$ and $b$ have order $1$. For bigger orders $m$ it seems easier to compute the number of pairs $(a,b)$ that have order that is a divisor of $m$ (instead of exactly $m$). For example, $(a,b)$ has order dividing $2$ (i.e. equal to $1$ or $2$) iff both $a$ and $b$ have order dividing $2$ (i.e. equal to $1$ or $2$). Thus we conclude $$\begin{align} \gamma_G(1)&=&\gamma_C(1)^2&=&1\\ \gamma_G(2)+\gamma_G(1)&=&(\gamma_C(2)+\gamma_C(1))^2&=&4\\ \gamma_G(3)+\gamma_G(1)&=&(\gamma_C(3)+\gamma_C(1))^2&=&9\\ \gamma_G(6)+\gamma_G(3)+\gamma_G(2)+\gamma_G(1)&=&(\gamma_C(6)+\gamma_C(3)+\gamma_C(2)+\gamma_C(1))^2&=&36&.\end{align}$$ Solving these equations, we obtain therefore $\gamma_G(1)=1$, $\gamma_G(2)=3$, $\gamma_G(3)=8$, $\gamma_G(6)=24$.