Suppose $f(z) = z + a_2 z^2 + \cdots$ with $f$ univalent on the unit disc, $f(0) = 0$, and $f'(0) = 1$. Define $g(z)^2 = f(z)/z$ and $h(z) = zg(z^2)$. Then $h$ is univalent on the disc, with $h(0) = 0$ and $h'(0) = 1$. Note that $h$ is odd, and $h(z)^2 = f(z^2)$ for each $z$ in the unit disc. How do you see that $h(z) = z + \frac{a_2}{2} z^2 + \cdots$?
EDIT: Added missing hypotheses.
On
I would suggest to just hold your nose and compute purely formally before doing anything else, using the standard expansion $\sqrt{1 + z} = 1 + \frac{1}{2}z - \frac{1}{8}z^2 + O(z^3)$. Since you're reading on Bieberbach's conjecture, I assume that you know how to do that and that you'll find the right answer.
If you believe that such an odd $h$ exists then make the ansatz as in Robert's answer or in my step 3 below and you should find the answer easily.
After having convinced ourselves that everything should work out well, I'd prove existence of $h$ as follows:
Step 1. Let $f: G \to \mathbb{C}$ be holomorphic with $0 \notin f(G)$ and assume that $G$ is a simply connected domain. Then $f$ has a holomorphic square root $h: G \to \mathbb{C}$, that is $f(z) = h(z)^2$ for $z \in G$.
Fix $z_0 \in G$ arbitrarily, choose $g(z_0)$ in such a way that $f(z_0) = \exp{g(z_0)}$ and put $g(z) = g(z_0) + \int_{z_0}^{z} \frac{f'(w)}{f(w)}\,dw$. Check that $g$ is well-defined, holomorphic, and that $\frac{d}{dz} [f \cdot e^{-g}] = 0$, hence $f = e^g$. Now put $h = e^{g/2}$.
[Note that $g(z) = \log{|f(z)|} + i \arg{f(z)}$ so that $g$ is unique up to addition of $2\pi i k$.]
Step 2. Let $f: \mathbb{D} \to \mathbb{C}$ be normalized schlicht, that is to say, $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ and $f$ is injective. There is an odd $h$ such that $f(z^2) = h(z)^2$. This $h$ is itself normalized schlicht.
Note that $\frac{f(z)}{z} = 1 + a_2 z + a_3 z^2 + \cdots$ is nowhere zero on $\mathbb{D}$ so that $\frac{f(z)}{z} = g(z)^2$ for some $g$ by step 1. Since $z \mapsto g(z^2)$ is even, the function $h(z) = z g(z^2)$ is odd and $h(z)^2 = f(z^2)$ by construction. Moreover, $h(0) =0$ and $h'(0) = g(0) = 1$.
If $h(z_1) = h(z_2)$ then $f(z_{1}^2) = f(z_{2}^2)$, hence $z_{1}^2 = z_{2}^2$ since $f$ is injective on $\mathbb{D}$, so either $z_1 = z_2$ or $z_1 = -z_2$; the latter case leads to $h(z_1) = h(-z_2) = -h(z_2)$ because $h$ is odd, so $h(z_1) = 0$. Since $f$ is schlicht, this implies that $z_1 = z_2 = 0$.
Step 3. If $f(z) = z + a_2 z^2 + \cdots $ is normalized schlicht, the function $h$ from step $2$ is of the form $h(z) = z + \frac{a_2}{2}z^3 + \cdots$.
Since $h$ is odd we may write $h(z) = z + b_3 z^3 + b_5 z^5 + \cdots$. From the relation $f(z^2) = h(z)^2$ we get $$\begin{align*} z^2 + a_2 z^4 + \cdots & = (z + b_3 z^3 + \cdots )(z + b_3 z^3 + \cdots ) \\ & = z^2 + 2b_3 z^4 + \cdots \end{align*}$$ and thus $b_3 = a_2/2$. Note that Robert gives $b_5 = \frac{a_3}{2} - \frac{a_2^2}{8}$ which you should try to verify yourself (that's also what you should get when doing the formal verification I suggested at the beginning of this post).
By the way: the map $f \mapsto h$ is called the square root transform and often one writes $h(z) = f(z^2)^{1/2}$. Similarly, we can define higher order transforms $h(z) = f(z^m)^{1/m}$ and these preserve normalized schlicht functions. One can even show that an odd normalized schlicht function is the square root transform of a normalized schlicht function.
It's actually not right. You could have e.g. $f(z) = z$, $h(z) = g(\arg z) z$ where $g$ takes values $\pm 1$ and $g(t) = g(\pi + t)$. But if you want $h$ to be an odd analytic function, it can't have any even powers of $z$ in its Maclaurin series: $h(z) = c_1 z + c_3 z^3 + \ldots$. Now $h(z)^2 = c_1^2 z^2 + 2 c_1 c_3 z^4 + \ldots$ while $f(z^2) = z^2 + a_2 z^4 + \ldots$, so $c_1^2 = 1$ and $a_2 = 2 c_1 c_3$. Thus either $c_1 = 1$ and $c_3 = a_2/2$ or $c_1 = -1$ and $c_3 = -a_2/2$. The next term in $h(z)$, by the way, in the case $c_1 = 1$ is $\left(\frac{a_3}{2} - \frac{a_2^2}{8}\right) z^5$.