Let $X,Y$ be topological spaces. It is typical exercise to prove $f:X\to Y$ is continuous iff $f(\bar A) \subset \overline{f(A)}$ (for any subset of the domain of $f$)
So I thought $f(A)^o\subset f(A^o)$ or $f(A^o)\subset f(A)^o$ can be another equivalent condition for continuity.($A^o$ is largest open set contained in $A$)
However, $f(A)^o\subset f(A^o)$ can't be true. $$f(x)=|x|,\quad A=\{x<0:x\text { is rational}\} \cup \{x>0 : x\text{ is irrational}\}$$ then $A^o=\emptyset$ but $f(A)=(0,\infty)$
In the case of $f(A^o)\subset f(A)^o$, I think this is true if $f$ is open mapping. But I'm not sure being open mapping is necessary.
You are right: for a function $f\colon X\longrightarrow Y$, $f$ being continuous is not equivalent to$$(\forall A\subset X):\mathring{\overbrace{f(A)}}\subset f\left(\mathring A\right).$$However, $f$ is continuous if and only if$$(\forall A\subset Y):f^{-1}\left(\mathring A\right)\subset\mathring{\overbrace{f^{-1}(A)}}.$$