The uniform continuity property that you have mentioned does guarantee that $X$ is a retract of the hyperspace $H(X)$. Let $[X]^{<\omega}$ denote the collection of all finite subsets of $X$. Then $[X]^{<\omega}$ is a dense subspace of $H(X)$. Suppose that for each $\epsilon>0$ there is a $\delta>0$ such that for each $A,B\subseteq X$ with $d_{H}(A,B)<\delta$, we have $d(f_{n}(A),f_{m}(B))<\epsilon$. Define a mapping $j:[X]^{<\omega}\rightarrow X$ by letting $j(A)=f_{n}(A)$ whenever $|A|=n$. Then $j$ is uniformly continuous.
It is well known that if $(X,\mathcal{U}),(Y,\mathcal{V})$ are uniform spaces and $(Y,\mathcal{V})$ is complete, then every uniformly continuous mapping $f:X\rightarrow Y$ extends to a uniformly continuous mapping $\overline{f}:\overline{X}\rightarrow Y$ where $\overline{X}$ is the completion of $\overline{X}$. Therefore, the mapping $j:[X]^{<\omega}\rightarrow $ extends to a uniformly continuous map $k:H(X)\rightarrow X$ and the mapping $k$ is clearly a retraction.
However, I must mention that it is often the case that there does not exist a retraction from $H(X)$ to $X$ since if $X$ is a Peano continuum and there is a retraction from $H(X)$ to $X$, then $X$ is contractible.
$\mathbf{Proposition}$ If $X$ is a Peano continuum and there is a retraction from $H(X)$ to $X$, then there is a deformation retract from $X$ to any point $x_{0}\in X$.
$\mathbf{Proof}$ By this answer, the hyperspace of any Peano continuum (locally connected continuum) is homeomorphic to the Hilbert cube. Since $H(X)$ is homeomorphic to the Hilbert cube, there is a deformation retract $T:H(X)\times[0,1]\rightarrow H(X)$ from $H(X)$ to the point $x_{0}$. In other words, $T$ is a continuous mapping with $T(x_{0},t)=x_{0}$ and $T(C,0)=x,T(C,1)=x_{0}$ for each $C\in H(X)$. Let $j:H(X)\rightarrow X$ be a retraction. Define a mapping $V:X\times[0,1]\rightarrow X$ by $V(x,t)=j T(x,t)$. Then $V$ is clearly a deformation retract from $X$ to the point $x_{0}$. $\mathbf{QED}$
$\textbf{There are sometimes retracts from spaces of measures}$
For this section, we shall suppose that $X$ is a compact metric space.
Using a mean to obtain a retraction from the hyperspace $H(X)$ to $X$ seems a little bit unnatural and the uniform continuity condition that you have mentioned is a bit unnatural as well (I must mention that I can imagine that there are contexts in which this uniform continuity condition would appear). We shall therefore now take a retraction from the space $\mathcal{M}(X)$ of all regular Borel probability measures on $X$ to $X$. The space $\mathcal{M}(X)$ shall be endowed with the weak$^{*}$-topology.
Let $\delta_{x}$ be the measure where $\int fd\delta_{x}=f(x)$ for each continuous $f$. By associating each point $x\in X$ with the measure $\delta_{x}$, we shall regard the space $X$ as a closed subspace of $\mathcal{M}(X)$. Take note that since $\mathcal{M}(X)$ is a convex subset of the space of all signed measures on $X$, there for each $x_{0}\in X$, there is a deformation retract of $\mathcal{M}(X)$ to any point measure $\delta_{x_{0}}$. Therefore, if there is a retract from $\mathcal{M}(X)$ to $X$, then for each $x_{0}\in X$, there is a deformation retract from $X$ to $x_{0}$.
Suppose now that $(X,d)$ is a compact metric space. The we shall call a family of functions $\mathcal{A}\subseteq\bigcup_{n\in\omega}C(X^{n},X)$ equicontinuous in the mean if for each $\epsilon>0$ there is some $\delta>0$ such that if $f\in\mathcal{A},f:X^{n}\rightarrow X$ and
if $\frac{1}{n}[d(x_{1},y_{1})+...+d(x_{n},y_{n})]<\delta$, then
$d(f(x_{1},...,x_{n}),f(y_{1},...,y_{n}))<\epsilon$.
We say that a continuous binary operation $*$ on $X$ is a mean if it satisfies the following identities
$x*x=x$
$x*y=y*x$
$(w*x)*(y*z)=(w*y)*(x*z)$.
Suppose that $*$ is a mean. Then define operations $m_{n}^{*}:X^{2^{n}}\rightarrow X$ (the operations $m_{n}^{*}$ were mentioned by Eric Wofsey in this answer) for all $n>0$ by induction:
$m^{*}_{1}(x,y)=x*y$
$m^{*}_{n+1}(x_{1},...,x_{2^{n+1}})=m^{*}_{n}(x_{1},...,x_{2^{n}})*m^{*}_{n}(x_{1},...,x_{2^{n}})$.
$\mathbf{Example}$ Suppose that $*$ is a mean on a compact metric space $(X,d)$ such that $d(x_{1}*x_{2},y_{1}*y_{2})<\frac{1}{2}(d(x_{1},y_{1})+d(x_{2},y_{2}))$. Then the family of mappings $\{m_{n}^{*}|n\in\mathbb{N}\}$ is equicontinuous in the mean.
Suppose now that $*$ is a mean on a compact metric space $(X,d)$.
Let $\mathcal{A}\subseteq\mathcal{M}(X)$ be the set of all measures of the form $\frac{1}{2^{n}}(\delta_{x_{1}}+...+\delta_{x_{2^{n}}})$. Since $\mathcal{M}(X)$ is a compact space in the weak$^{*}$-topology, the space $\mathcal{M}(X)$ has a unique uniformity.
Suppose that $*$ is a mean on a compact metric space $(X,d)$. Define a mapping $j_{*}:\mathcal{A}\rightarrow X$ by letting $j_{*}(\frac{1}{2^{n}}(\delta_{x_{1}}+...+\delta_{x_{2^{n}}}))=m^{*}(x_{1},...,x_{2^{n}})$.
$\mathbf{Theorem}$ Suppose that $*$ is a mean on a compact metric space such that $\{m^{*}_{n})|n\in\mathbb{N}\}$ is equicontiuous in the mean. Then the mapping $j_{*}$ is uniformly continuous. In particular, the mapping $j_{*}$ extends to a unique mapping $k_{*}:\mathcal{M}(X)\rightarrow X$ (and the mapping $k_{*}$ is a retract).
$\mathbf{Proof}$ For this theorem, let $D=\max_{x,y\in X}d(x,y)$
Suppose that $\epsilon>0$. Then since $\{m_{n}^{*}|n\in\mathbb{N}\}$ is equicontinuous in the mean, there is some $\delta>0$ such that if $n>0$ and $\frac{1}{2^{n}}(d(x_{1},y_{1})+...+d(x_{2^{n}},y_{2^{n}}))<\delta$, then $d(m_{n}^{*}(x_{1},...,x_{2^{n}}),m_{n}^{*}(y_{1},...,y_{2^{n}})<\epsilon$
Now let $\beta>0,\alpha>0$. Then there are $f_{1},...,f_{k}:X\rightarrow[0,1]$ such that whenever $1\leq i\leq k$ the support $\textrm{supp}(f_{i})$ has diameter less than $\alpha$ and where
$\{(f_{i}^{-1}[\{1\}])^{\circ}|1\leq i\leq k\}$ covers $X$. Let $U_{i}=f^{-1}_{i}[\{1\}]^{\circ}$ whenever $1\leq i\leq k$.
If $A\subseteq\{1,...,k\}$, then let $f_{A}=\max_{a\in A}f_{a}$.
Now suppose that $\mu,\nu\in\mathcal{A}$ and
$$|\int f_{A}d\mu-\int f_{A}d\nu|<\beta$$ for each $A\subseteq\{1,...,k\}$.
Then there is some $n$ and $x_{1},...,x_{2^{n}},y_{1},...,y_{2^{n}}$ such that
$$\mu=\frac{1}{2^{n}}(\delta_{x_{1}}+...+\delta_{x_{2^{n}}}),\nu=\frac{1}{2^{n}}(\delta_{y_{1}}+...+\delta_{y_{2^{n}}})$$. Then since $$\int f_{A}d\mu=\frac{1}{2^{n}}(f_{A}(x_{1})+...+f_{A}(x_{2^{n}})),\int f_{A}d\nu=\frac{1}{2^{n}}(f_{A}(y_{1})+...+f_{A}(y_{2^{n}})),$$ we have
$$|\frac{1}{2^{n}}(f_{A}(x_{1})+...+f_{A}(x_{2^{n}}))-\frac{1}{2^{n}}(f_{A}(y_{1})+...+f_{A}(x_{2^{n}})|<\beta$$
for each $A\subseteq\{1,...,k\}$, so
$|(f_{A}(x_{1})+...+f_{A}(x_{2^{n}}))-(f_{A}(y_{1})+...+f_{A}(y_{2^{n}}))|<2^{n}\beta$.
Let $r=\lfloor 2^{n}\cdot\beta\rfloor$.
If $1\leq i\leq 2^{n}$, then let $M(i)=\{j\in\{1,...,2^{n}\}|d(x_{i},y_{j})<\alpha\}$ and if $R\subseteq\{1,...,2^{n}\}$, then let $M(R)=\bigcup_{r\in R}M(r)$.
Suppose now that $R\subseteq\{1,...,2^{n}\}$. Let $a_{r}\in\{1,...,k\}$ be such that $x_{r}\in U_{a_{r}}$. Let $A=\{a_{r}|r\in R\}$.
Then $f_{A}(x_{r})=1$ for each $r\in R$, so
$f_{A}(x_{1})+...+f_{A}(x_{2^{n}})\geq |R|$. Therefore, $f_{A}(y_{1})+...+f_{A}(y_{2^{n}})>|R|-2^{n}\beta$.
Take note that there must be at least $|R|-r$ indices $i$ such that $f_{A}(y_{i})\neq 0$.
However, if $f_{A}(y_{i})>0$, the $y_{i}\in\mathrm{supp}(f_{a_{r}})$ for some $r\in R$, so since $\mathbf{supp}(f_{a_{r}})$ has diameter less than $\alpha$, we conclude that $y_{i}$ is within $\alpha$ distance of $x_{r}$. We therefore conclude that $i\in M(r)$, so $i\in M(R)$. Therefore, we conclude that $|M(R)|\geq|R|-r$.
If $1\leq i\leq 2^{n}$, then let $N(i)=\{j:1\leq j\leq 2^{n},d(x_{i},y_{i})<\alpha\}\cup\{2^{n}+1,...,2^{n}+r\}$ and if $2^{n}<i\leq 2^{n}+r$, then let $N(i)=\{1,...,2^{n}+r\}$. If $A\subseteq\{1,...,2^{n}+r\}$, then define $N(A)=\bigcup_{a\in A}N(a)$. Then $|N(A)|\geq|A|$ for each $A\subseteq\{1,...,2^{n}+r\}$. Therefore, by Hall's marriage theorem, there exists a bijection $f:\{1,...,2^{n}+r\}\rightarrow\{1,...,2^{n}+r\}$ such that $f(i)\in N(i)$ for $1\leq i\leq 2^{n}+r$. Therefore, by restricting $f$ to a function whose domain and range are subsets of $\{1,...,2^{n}\}$, we conclude that there are $A,B\subseteq\{1,...,2^{n}\}$ such that $|A|\geq 2^{n}-r,|B|\geq 2^{n}-r$ and a bijection $h:A\rightarrow B$ such that $h(i)\in N(i)$ for $i\in A$. Extend the bijection $h$ to a bijection $g:\{1,...,2^{k}\}\rightarrow\{1,...,2^{k}\}$. Then
$$\frac{1}{2^{n}}(d(x_{1},y_{g(1)})+...+d(x_{2^{n}},y_{g(2^{n})}))$$
$$=\frac{1}{2^{n}}[\sum_{a\in A}d(x_{a},y_{g(a)})+\sum_{b\in\{1,...,2^{n}\}\setminus A}d(x_{b},y_{g(b)})]$$
$$<\frac{1}{2^{n}}(2^{n}\cdot\alpha+r D)=\alpha+D\frac{r}{2^{n}}\leq \alpha+D\beta.$$
Therefore, choose $\alpha,\beta$ such that $\alpha+D\beta<\delta$. Then we have
$$\frac{1}{2^{n}}(d(x_{1},y_{g(1)})+...+d(x_{2^{n}},y_{g(2^{n})}))<\delta.$$
Therefore, we have
$$d(j_{*}(\mu),j_{*}(\nu))=d(m^{*}_{n}(x_{1},...,x_{2^{n}}),m^{*}_{n}(y_{1},...,y_{2^{n}})))$$
$$=d(m^{*}_{n}(x_{1},...,x_{2^{n}}),m^{*}_{n}(y_{g(1)},...,y_{g(2^{n})}))<\epsilon.$$
We conclude that the mapping $j^{*}$ is uniformly continuous.
$\mathbf{QED}$
