So if we have the infinitesimal generator of a Markov process $\{X_t\}_{t\geq0}$, the infinitesimal generator is given by $$Af(x) = \lim_{t\rightarrow 0} \frac{1}{t} \big(P_tf(x) - f(x)\big)$$ where $P_tf(x) = \mathbb{E}[f(X_t)|X_0 = x]$. Then the Kolmogorov forward equation is the PDE
$$KFE: \begin{cases}\frac{\partial }{\partial t}u(t,x) &= Au(t,x)\\ u(0,x) &= f(x) \end{cases}$$
which is satisfied by $u(t,x) = \mathbb{E}[f(X_t)|X_0 = x]$.
My understanding is that $Af(x)$ is like the "expected derivative" wrt $t$ of $f(X_t)$, i.e. the expected infinitesimal change in $f(X_t)$.
So why is KFE interesting or important? Doesn't it just restate that $AP_tf(x)$ is the time derivative of $P_tf(x)$ which is true by definition? And what exactly is the implication?
Also: I understand that for diffusion processes like Brownian motion, $A$ resembles the Laplacian operator so KFE is the heat equation. Intuitively, diffusion with an initial condition of $f(x)$ would basically flatten out $f(x)$, so this means the expected value of $f(B_t)$ decreases over time. But for symmetric Brownian motion, shouldn't $P_tf(x) = f(x)$ for all $t$?