There are two separate issues that need to be addressed. First, how to understand the minimization of the Helmholtz energy; and second, how the Helmholtz energy relates to work. I will address each separately.
Minimization of $A(T,V,n)$
We consider a system with uniform temperature $T$, fixed total volume $V$ and fixed number of particle $n_i$ of type $i=1,2\cdots K$ (the system may be multicomponent). At equilibrium the Helmhotlz energy of this system is
$$A(T, V_A+V_B, n_{Ai}+n_{Bi}) \tag{1}$$
This system has uniform temperature $T$, uniform pressure $P$ and uniform chemical potential $\mu_{i}$ for all components.
Next we divide this system arbitrarily into two parts: part $A$ whose state is $(T,V_A, n_{Ai})$, and part $B$ whose state is $(T,V_B=V-V_A, n_{Bi}=n_i-n_{Ai}$. Each part is in internal equilibrium, i.e., compartment $A$ has uniform $T$, $P_A$, $\mu_{Ai}$, and compartment $B$ has uniform $T$, $P_B$, $\mu_{Bi}$.
For arbitrary partitioning, $P_A\neq P_B$ and $\mu_{Ai}\neq \mu_{iB}$, which means that generally the parts are not in equilibrium with each other. As a consequence of the second law we must have
$$
\tag{2}
A(T, V_A, n_{Ai}) + A(T, V_B, n_{Bi}) \geq A(T, V_A+V_B, n_{Ai}+n_{Bi})
$$
and in more condensed form
$$
\tag{3}
A_A + A_B - A \geq 0 \Rightarrow \Delta A = A - (A_A+A_B) \leq 0
$$
The equal sign applies if and only if $P_{A}=P_{B}$ and $\mu_{Ai}=\mu_{Bi}$, i.e., when the parts are in equilibrium with respect to each other. In other words:
the equilibrium state has the lowest Helmholtz energy among all possible ways to partition $V$ and $n_i$ into any number of parts under uniform temperature
Notice that in this construction the parts are always in internal equilibrium. This is the way that equilibrium thermodynamics treats non equilibrium states, in this case a gradient in pressure and chemical potential.
Also notice that as we partition $V$ and $n_i$ at fixed $T$ the internal energy of the system will vary between different partitions. We assume the presence of a bath that can provide or absorb the required energy. This is the bath that ensures $T$ is the same in all parts at all times.
Helmholtz energy and work
The inherent tendency of systems to reach equilibrium is associated with the ability to produce work. A non equilibrium state is characterized by gradients which could be leveraged to produce work. This is a very general principle: a waterfall, a voltage difference, a temperature difference, all of such gradients can produce work – but only if we place a suitable machine in the path of the flow (of fluid, current, heat, etc). The difference of the Helmholtz energy between the final (equilibrium) state and the initial (non-equilibrium) state is the maximum amount of work that can be extracted:
$$\tag{4}
W = A - (A_A + A_B) = \Delta A \leq 0
$$
I want to correct a statement in the original post:
According to the inequality $W_{by}\le-\Delta F$, as the system does work on surroundings, its $F$ decreases. The system will keep doing work until its $F$ reaches the minimum value which only happens when the system is in thermal equilibrium with the surroundings. In this state, the system is unable to do any further work.
This is not true. Since the total volume is fixed, no work is done against the surroundings. To obtain work from the equilibration process we need to build some machine with weights, springs, semipermeable membranes and similar mechanical contraptions. If no such machinery is present the amount of work is zero, just like a waterfall with no turbine in its path to capture its energy before it is converted into internal energy at the bottom.
