Yes.
If we have a space $X$ and an equivelence relation $\sim$ on $X$, then the standard approach is to endow the set $X/\sim$ of equivalence classes with respect to $\sim$ with the quotient topology induced by the canonical map $q : X \to X/\sim$ given by $q(x) =[x]$ = equivalence class of $x$.
Therefore, whenever such a set $X/\sim$ occurs in topology, you can be sure that it is automatically regarded as a topological space with the quotient topology.
In your example only the quotient topology can assure the continuity of the map $\tilde f : \mathbb R/\mathbb Z \to S^1$ which is induced by $f$.
Let me mention that the notation $\mathbb R/\mathbb Z$ is somewhat ambiguous. In topology, if we are given a space $X$ and a subset $A \subset X$, then $X/A$ usually denotes the set obtained by collapsing $A$ to a point. Formally we have $X/A = X/\sim$ with $x \sim y$ iff $x = y$ or $x,y \in A$. But this is not the interpretation of $\mathbb R/\mathbb Z$. Here it is regarded as the quotient group of the additive group $(\mathbb R, +)$ by its subgroup $\mathbb Z$.
See also $\mathbb{R} / \mathbb{Z}$ is homeomorphic to $S^{1}$ .