= Solution
Use the curvature-minus-one normalization of the <hyperbolic metric>:
$$
\boxed{ds_{\mathbb D}=\lambda(z)|dz|,\qquad\lambda(z)=\frac2{1-|z|^2}.}
$$
The associated distance is $d(z,w)=2\operatorname{artanh}|(z-w)/(1-\overline wz)|$. A connected <Riemann surface> has its complete canonical <Poincare metric on a Riemann surface> precisely when its <universal cover> is conformally the <unit disc>. By the <uniformization theorem>, the other <simply connected> covering types are the complex plane and <Riemann sphere>. Disc automorphisms preserve $\lambda(z)|dz|$, as direct differentiation of $(z-a)/(1-\overline az)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.
First prove the needed <Schwarz lemma>. If $F:\mathbb D\to\mathbb D$ is <holomorphic> and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The <maximum modulus principle> and $r\uparrow1$ give $|F(z)|\le|z|$ and $|F'(0)|\le1$.
Write $\psi_a(z)=(z-a)/(1-\overline az)$. Conjugating $f$ by the source and target disc automorphisms, $\psi_{f(a)}\circ f\circ\psi_a^{-1}$ fixes zero. Applying the <derivative> bound gives
$$
\boxed{\frac{|f'(a)|}{1-|f(a)|^2}\le\frac1{1-|a|^2},\qquad f^*ds_{\mathbb D}\le ds_{\mathbb D}.}
$$
Integrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the <Schwarz-Pick theorem>. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.
If both $R$ and $S$ have disc universal covers, lift $g\circ\pi_R$ to $\widetilde g:\mathbb D\to\mathbb D$ through $\pi_S$. The disc is <simply connected>, so the lift exists, and it is <holomorphic> because the <covering maps> are local <biholomorphisms>. Both <covering maps> are local hyperbolic isometries. The disc contraction therefore descends to $g^*ds_S\le ds_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a <hyperbolic metric>, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere <universal cover> of $R$ to the disc cover of $S$. A bounded entire function is constant by the <Liouville theorem>, and a <holomorphic map> from the compact sphere into the disc is constant by the <maximum modulus principle>. Thus \b[a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps].
For the prescribed point, <Schwarz-Pick theorem> gives $|\psi_{w_0}(f(z))|\le|\psi_{z_0}(z)|$. Therefore
$$
h(z)=\frac{\psi_{w_0}(f(z))}{\psi_{z_0}(z)}
$$
is <holomorphic> after filling the <removable singularity> at $z_0$, and $|h|\le1$. Inverting the target automorphism proves
$$
\boxed{f(z)=\frac{(z-z_0)h(z)+w_0(1-\overline z_0z)}{\overline w_0(z-z_0)h(z)+(1-\overline z_0z)},\qquad\sup_{\mathbb D}|h|\le1.}
$$
Conversely every <holomorphic> $h$ with this bound gives an admissible disc map: $|\psi_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.
Fix $z$ and put $r=|\psi_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le1$, attain them all. Thus the <one-point value region of a holomorphic disc map> is $|\psi_{w_0}(w)|\le r$. Completing the square in $|w-w_0|^2\le r^2|1-\overline w_0w|^2$ gives its Euclidean centre and radius:
$$
\boxed{C=\frac{(1-r^2)w_0}{1-r^2|w_0|^2},\qquad s=\frac{r(1-|w_0|^2)}{1-r^2|w_0|^2}.}
$$
This closed disc lies strictly inside the <unit disc> because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.
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