I'm slightly puzzled by the following: if $g(t)$ is a function in $L^q(X)$ then we can show that $g(t-x)$ is continuous function of $t$, i.e. for $\varepsilon > 0$ we can find $\delta$ such that $d(x,y)<\delta$ implies $\|g(t-x) - g(t-y)\|_q < \varepsilon$.
But $g$ is not necessarily continuous. Is this result stating something like $g$ is continuous with respect to norm $\|\cdot\|_q$? Because if $\tau_x$ is translation by $x$ then $g(t-x) = g \circ \tau_x$ is not necessarily continuous. When continuous I mean in topology on $X$ (e.g. $X \to \mathbb R$). Does this sort of continuity have a name?
You are confusing things. First we need to assume that $X$ is a subset of a vector space $V$ so we can form differences. Then we need to assume that $g\circ \tau_t$ is defined on $X$ if $g$ is, e.g. by assuming that $X = X - t$ (e.g. if $X=\mathbb{R}$) or by assuming that $g$ has compact support, whatever -- there are several possibilities. What is meant by the continuity of translation is that the map $t\mapsto g\circ\tau_t $ is continuous as a map from a neighbourhood of $0\in V$ to $L^q(X)$.
This does not imply in any sense that the translated function is continuous, but it means that if $|s-t|$ is small, then $||g\circ\tau_s - g\circ\tau_t||_q$ is small, too (in the $\varepsilon, \delta$ sense).