Let $(X_t)$ be a sample-continuous stochastic process and $f$ a real measurable bounded function.
Let $(\mathcal F_t)$ be the filtration generated by $(X_t)$ and $(\mathcal G_t)$ the filtration generated by $(f(X_t))$.
Let $0<s<t$. By the tower property of conditional expectation, we have
$$E[f(X_t)| \mathcal G_s] = E[E[f(X_t)| \mathcal F_s] | \mathcal G_s]=E[E[f(X_t)| \mathcal G_s] | \mathcal F_s].$$
Now suppose that $(X_t)$ is Markov w.r.t its own filtration, i.e, $E[f(X_t)| \mathcal F_s]=E[f(X_t)| X_s]$. This means that $$E[f(X_t)| \mathcal G_s]=E[E[f(X_t)| X_s] | \mathcal G_s]=E[E[f(X_t)| \mathcal G_s] | X_s]$$
First question: does this mean that $\mathcal G_s \subseteq \sigma(X_s)$ ? (this would be some sort of converse to the tower property)
Second question: does this mean that $E[E[f(X_t)| X_s] | \mathcal G_s]=E[E[f(X_t)| X_s] | \sigma(f(X_s))] $?
The $\sigma$-algebra ${\cal G}_s$ is generated by the sets $\{f(X_t)\in B\}$ where $t\le s$ and $B\in {\cal B}(\mathbb R)\,.$ Because every such set can be written as $\{X_t\in f^{-1}(B)\}$ it is in $\sigma(X_t)\,.$ Since this holds for all $t\le s$ this shows $$ {\cal G}_s\subseteq{\cal F}_s\,. $$ It is not necessarily true that ${\cal G}_s\subseteq\sigma(X_s)$ which is smaller than ${\cal F}_s\,.$ The tower property of conditional expectations is not needed to examine such relationships.
The answer to your last question is no because ${\cal G}_s$ is generated not only by $f(X_s)$ but by all $f(X_t)$ with $t\le s\,.$
A related question to yours is this one. It has no answer but at least some reference.