I tried to prove the statement for a while and I think I managed to narrow it down to one particularly difficult case. In the end, it led me to a counter example, showing there are no such values $g$ and $t$. This came as a bit of a surprise for me. The construction goes as follows.

(1) For every $n \geq 1$ there is a cycle $C$ and a labeling $\varphi: V(C) \to [n+1]$ such that $|\varphi^{-1}(n+1)| = 1$ and for every non-trivial path $P = xPy \subseteq C$ and all $i < \min\{\varphi(x), \varphi(y)\}$, $P$ contains a vertex labeled $i$.

proof: By induction on $n$, the case $n =1$ being trivial. In the inductive step, start from $(C, \varphi)$ for $n$, and obtain $C'$ from $C$ by subdividing every edge. Let $\varphi'(x) = \varphi(x)+1$ for $x \in C$ and $\varphi'(x) = 1$ for $x \in C' \setminus C$.

(2) Let now $n$ be given. Start with the disjoin union of $n$ copies $C_1, \ldots, C_n$ of the labled cycle from (1). Subdivide every edge of each cycle $n$ times, leaving the new vertices unlabeled. For every $i$, let $x_i \in C_i$ be the unique vertex labeled $n+1$. Join $x_i$ to all vertices on $\bigcup_{i < j \leq n} C_j$ labeled $i$.

It's easy to see that every cycle $D$ must contain at least one of $x_1, \ldots, x_n$. Let the minimum $1 \leq i \leq n$ with $x_i \in D$ be the index $\mathcal{idx}(D)$ of $D$. Moreover, we can see that $D$ contains a neighbor of $x_i$ for all $i < \mathcal{idx}(D)$.

Let $D_1, D_2$ be two cycles of $G$, wlog $\mathcal{idx}(D_1) \leq \mathcal{idx}(D_2)$. If equality holds, then $D_1 \cap D_2$ is non-empty. If $\mathcal{idx}(D_1) <\mathcal{idx}(D_2)$, then there is an edge from $D_1$ to $D_2$. Either way, any two cycles touch.

Moreover, since $G$ has disjoint pairwise touching cycles $C_1, \ldots , C_n$, the tree-width of $G$ is at least $n-1$. Since every cycle must contain an edge of at least one cycle $C_i$, the girth of $G$ is at least $n$.