If $h$ is uniformly continuous on $\mathbb{R}$, then $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\forall x,y \in dom(h)$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.
My proof:
Let $\epsilon > 0$. Since $g$ is uniformly continuous, $\exists \delta > 0$ s.t. $\forall x,y \in \mathbb{R}$, $|x - y| < \delta$ implies $|g(x) - g(y)| < \epsilon$.
Now, $f$ uniformly continuous means that $\exists \delta' > 0$ s.t. $\forall x',y' \in \mathbb{R}$, $|x' - y'| < \delta'$ implies $|f(x') - f(y')|< \delta$.
So, for any $x', y' \in \mathbb{R}$ with $|x' - y'| < \delta'$, we have $|f(x') - f(y')|< \delta$. Hence $|g(f(x')) - g(f(y'))| < \epsilon$. This is exactly what it means for $g \circ f: \mathbb{R} \rightarrow \mathbb{R}$ to be uniformly continuous, so we are done.
Thank you for your feedback.
To prove that $g\circ f$ is uniformly continuous on $\mathbb{R},$ one needs to prove that for any $\varepsilon>0$, there exists $\delta>0$ such that for all $x,y\in \mathbb{R},$ if $|x-y|<\delta,$ then $$|g(f(x)) - g(f(y))|<\varepsilon.$$
Now, fix $\varepsilon>0.$ Since $g$ is uniformly continuous on $\mathbb{R},$ there exists $\eta>0$ such that for all $x,y\in \mathbb{R},$ if $|x-y|<\eta,$ then $$|g(x) - g(y)|<\varepsilon.$$ Since $f$ is uniformly continuous on $\mathbb{R},$ there exists $\delta>0$ such that for all $x,y\in\mathbb{R},$ if $|x-y|<\delta,$ then $$|f(x)-f(y)|<\eta.$$
We claim that such $\delta>0$ will work. Indeed, fix $x,y\in\mathbb{R}$ such that $|x-y|<\delta.$ By uniform continuity of $f$, we have
$$|f(x)-f(y)|<\eta.$$ By uniformly continuity of $g,$ we have $$|gf(x))-gf(y))|<\varepsilon.$$ This concludes that $g\circ f$ is uniformly continuous on $\mathbb{R}.$