Denote $$H=\{z=x+iy: x,y\in \mathbb{R}, y>0\}, \quad \text{and}\\[12pt]V=\{z=x+iy: x^2-y^2>1, x>0\}.$$
Let $$\mathcal{S} = \{f:H\to V, \,\,f \text{ is holomorphic and } f(i)=2\}.$$
Find all the functions $g\in \mathcal{S}$ such that $|f'(i)|\leq |g'(i)|$ for all $f\in \mathcal{S}$. Compute $g'(i)$.
Ideas: Suppose we have $\varphi: \mathbb{D}\to H:0\mapsto i$ and $\psi: V\to \mathbb{D}:2\mapsto 0$. Then we can reason by Schwarz's lemma somehow, but I'm not seeing a bound on $\{f'(i)\mid f\in H\}$.
Your idea is exactly right (assuming you mean $\psi$ and $\varphi$ to be conformal maps onto the corresponding domains). By the chain rule, the derivative of the composition $\psi\circ f\circ \varphi$ at $0$ is $$ \varphi'(0) \, f'(i)\, \psi'(2)$$ By the Schwarz lemma, this is at most $1$ in absolute value. Hence, $$|f'(i)|\le \frac{1}{|\varphi'(0)\psi'(2)|}\tag{1}$$ Equality is attained in (1) if and only if $\psi\circ f\circ \varphi$ is a rotation, which means there is real constant $\alpha$ such that $$f(z)= \psi^{-1}(e^{i\alpha} \varphi^{-1}(z)),\quad z\in H \tag{2}$$ This identifies the set of functions $f$ which maximize $|f'(i)|$. Of course, without concrete $\psi$ and $\varphi$ this is not sufficiently explicit. Also, we need $\psi$ and $\varphi$ to find the value of the maximum.
To find $\varphi$ is standard:
$$\varphi(z)=i\frac{1-z}{1+z}$$ As for $\psi$, note that $x^2-y^2$ is the real part of $z^2$. Thus, $z\mapsto z^2-1$ sends $V$ onto the right halfplane. From there it's standard: $$\psi(z) = \frac{(z^2-1)-3}{(z^2-1)+3}$$