An estimate on the coefficient of bounded schlicht functions

Concerning the first part: You can indeed modify the well-known proof of $|a_2| \le 2$. If $|f(z)| \le M$ in $\Bbb D$ then $g^*(z) \ge \frac{1}{\sqrt M}$ in $\Bbb C \setminus \Bbb D$. So the complement of the image of $g^*$ contains a disk with the area $\pi/M$, and Gronwall's area theorem gives $$ \pi \left( 1 - \sum_{n=1}^\infty n |b_n^*|^2 \right) \ge \frac{\pi}{M} \\ \implies \sum_{n=1}^\infty n |b_n^*|^2 \le 1 - \frac 1M \, . $$ In particular, $$ \frac{|a_2|}{2} = |b_1^*| \le 1 - \frac 1M \implies |a_2| \le 2 \left( 1 - \frac 1M\right) \, . $$

Another way that solves both problems at the same time is to consider $K(w)=\frac{2w+1-(1+4w)^\frac{1}{2}}{2w}$ the inverse of the Koebe function $k(z)=\frac{z}{(1-z)^2}$, where the square root is the principal one so it is $1$ for $w=0$ and $K$ of course is defined on the plane minus the usual half-line from $[-\frac{1}{4}, -\infty)$.

Then if $f$ is univalent normalized and $|f(z)| \le M$, we get that $Mk(\frac{f}{M})$ is normalized univalent, and writing that out we get that its second coefficient is $a_2+\frac{2}{M}$, hence we get $|a_2+\frac{2}{M}| \le 2$.

But we can rotate $f$ by the argument of $a_2$ and keep it with same properties ($g(z)=e^{it}f(ze^{-it}), a_2e^{-it}=|a_2|$), so applying the above we actually get $||a_2|+\frac{2}{M}| \le 2$ hence the required inequality on $|a_2|$

For part 2 we use that $f(z)=MK(\frac{g(z)}{M})$ for some normalized univalent $g$ and using that $g(\mathbb D)$ contains the disc of radius $\frac{1}{4}$, we immediately get that $(f/M)(\mathbb D)$ contains the disc obtained when we minimize the absolute value of the expression of $K$ on the disc of radius $\frac{1}{4M}$ and it is an easy calculation which shows that happens at $w=-\frac{1}{4M}$ and after some standard manipulations we get that $f(\mathbb D)$ contains the disc of radius $M(2M-1-2(M^2-M)^\frac{1}{2})=\frac{1}{2(1-\frac{1}{2M}+\sqrt{1-\frac1M})}$ and that contains the one needed in the OP claim