Let's consider an explicit example. Take $\beta=\omega$, and $\alpha=\omega\cup\{\omega\}$. Define $f\colon\alpha\to\beta$ as $f(n)=n+1$ if $n\in\omega$, and $f(\omega)=0$.
The direct image function $\overline{f}\colon\mathcal{P}(\alpha)\to\mathcal{P}(\beta)$ determined by $f$ from $\alpha$ to $\beta$ does not agree with $f$ itself: $f(\omega)=0$, but the direct image function has $\overline{f}(\{0,1,\ldots,\}) = \{1,2,3\ldots\}\neq 0$. In fact, the values of the direct image function are not elements of $\beta$: for instance, $\overline{f}(\{0\}) = \{1\}$, which is not an element of $\beta$ (since it is not an ordinal).
So the direct image function does not give a bijection between $\alpha$ and $\beta$: it doesn't even map $\alpha$ to $\beta$; so the fact that $x\subseteq y$ implies $\overline{f}(x) \subseteq \overline{f}(y)$ is not really relevant here: you are not using the direct image function. You are using your original $f$ on the elements of $\alpha$ and $\beta$, it's just that you are looking at $\alpha$ as ordered via $\subseteq$ instead of ordered via $\in$. And if $x,y\in\alpha$ are such that $x\lt y$ but $f(x)\not\lt f(y)$, then you have $x\subseteq y$ but you still don't have $f(x)$ (the element of $\beta$) contained in $f(y)$.