Let $G$ be a topological group, and $f: G \rightarrow \mathbb C$ function. We say $f$ is uniformly continuous if for any $\epsilon > 0$, there exists a neighborhood $W$ of $1_G$ such that $xy^{-1} \in W$ implies $|f(x) - f(y)| < \epsilon$.
We say $f$ is compactly supported if there is a compact set $\Omega \subset G$ such that $f(g) \neq 0$ implies $g \in G$.
Two questions:
(i): If $f$ is uniformly continuous, is it continuous in the ordinary sense?
(ii): If $f$ is compactly supported and continuous, is it uniformly continuous?
(i): Assume $f$ is uniformly continuous. Let $x \in G$ and $\epsilon > 0$ be given. To show that $f$ is continuous, we must show there is a neighborhood $U$ of $x$ such that $y \in U$ implies $|f(x) - f(y)| < \epsilon$. We know there is an open neighborhood $W$ of $1_G$ such that $yx^{-1} \in W$ implies $|f(x) - f(y)| < \epsilon$. Taking the neighborhood $U = Wx$ of $x$, we see that if $y \in U$, then $yx^{-1} \in W$, and therefore $|f(x) - f(y)| < \epsilon$ as required.
(ii): Assume $f$ is continuous and compactly supported. Given $\epsilon > 0$, we must find a neighborhood $W$ of $1_G$ such that $xy^{-1} \in W$ implies $|f(x) - f(y)|< \epsilon$.
Let $\Omega$ be a compact set outside of which $f$ vanishes. For each $x_i \in \Omega$, let $U_{x_i}$ be a neighborhood of $x_i$ for which $y \in U_{x_i}$ implies $|f(y) - f(x_i)| < \epsilon/2$. If we consider each neighborhood $U_{x_i}x_i^{-1}$ of $1_G$, a basic result about topological groups tells us there is a symmetric open neighborhood $W_{x_i}$ of $1_G$ (that is, $W_{x_i}^{-1} = W_{x_i}$) such that the product set $W_{x_i}W_{x_i}$ is contained in $U_{x_i}x_i^{-1}$.
The sets $W_{x_i}x_i$ cover $\Omega$, and admit a finite subcover $ W_{x_1}x_1 \cup \cdots \cup W_{x_t}x_t$.
Let $W = W_{x_1} \cap \cdots \cap W_{x_t}$. Then $W$ is an open neighborhood of $1_G$, and we claim that $|f(x) - f(y)| < \epsilon$ whenever $xy^{-1} \in W$.
Indeed, if neither $x$ nor $y$ lie in $\Omega$, then $f(x) = f(y) = 0$, and the claim is obvious. If on the other hand, $x \in \Omega$, then $x \in W_{x_i}x_i$ for some $1 \leq i \leq t$. Then $xx_i^{-1} \in W_{x_i}$, and we also have $yx^{-1} \in W_{x_i}$ (indeed, $yx^{-1} = (xy^{-1})^{-1} \in W^{-1} \subset W_{x_i}^{-1} = W_{x_i}$).
Then
$$yx_i^{-1} = (yx^{-1})(xx_i^{-1}) \in W_{x_i} W_{x_i} \subset U_{x_i}x_i^{-1}$$
and hence $y \in U_{x_i}$. So both $x$ and $y$ are in $U_{x_i}$, which means that
$$|f(x) - f(x_i)| < \epsilon/2$$
$$|f(y) - f(x_i)| < \epsilon/2.$$
Therefore, $|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(y) - f(x_i)| < \epsilon$.