There are two terms in the Van der Waals correction:
$$\left(P + a\left(\frac{n}{V}\right)^2\right)\left(\frac{V}{n}-b\right)=RT$$
The first of these ($a$) is an attraction term: the molecules attract each other, which makes the pressure lower. It's the sum of the actual pressure and the attraction term that gives you the "ideal gas pressure".
The second term ($b$) is a volume term: the "apparent volume" that the molecules can move in is smaller than the volume of the container, so we need to subtract a number from the size of the container to get the "ideal volume".
And this is why you have the two terms, with opposing signs.
Because the Helmholtz free energy graph (at a fixed temperature) is not convex, we can draw the following tangent line. It touches the graph at two points A and B, with corresponding volumes $V_A$ and $V_B$, and free energies $F_A$ and $F_B$. These two points are special, for reasons we'll get to in a moment.
Suppose our system has a volume that's exactly halfway between $V_A$ and $V_B$. Let's consider two possible states the system could be in:
A homogeneous state, with a constant density throughout the system. This density is determined by the van der Waals equation of state, and so its Helmholtz free energy lies on the purple curve (which is constructed from the equation of state directly.)
A heterogeneous state, where the system has divided itself into two parts: half of the volume is in state $A$, while half of the volume is in state $B$. The total volume of the system is halfway between $V_A$ and $V_B$; and the total Helmholtz free energy of the system is also halfway between $F_A$ and $F_B$. In other words, the total Helmholtz free energy of the system lies along the red tangent line, halfway between $A$ and $B$.
But for a fixed temperature and volume, the Helmholtz free energy is minimized in equilibrium. This means that the heterogeneous state must be the equilibrium state, since its $F$ is lower. A similar logic can be applied to any volume between $V_A$ and $V_B$; for example, if the volume of our system is 10% of the way from $V_A$ to $V_B$, then the equilibrium state will be a heterogenous state where 10% of the volume is in state A and 90% is in state B.
In other words, for any volume between $V_A$ and $V_B$, the system "wants" to separate itself into two parts of different densities, since that will give it a lower Helmholtz free energy. If we imagine compressing this system at fixed $T$ from a large initial volume, it would initially be in a less-dense state, with a $V$ and $F$ somewhere along the purple curve to the right. When we compress it down to $V_B$, we would start to see the system separate into two parts: one with the same density the system had at $V_B$, and another more dense part with higher density. This denser part of the system would grow until the volume decreased to $V_A$; during this transition, the system's total $V$ and $F$ lie somewhere along the red curve. Finally, when we reached $V_A$, the system would be entirely in the denser phase, and further compression would cause $V$ and $F$ to follow the purple curve on the left.
If you're paying close attention, you might notice that I've implicitly assumed that these two "subsystems" in states A and B are in equilibrium with each other. But is this actually the case? Thankfully, yes. The two subsystems can exchange energy, volume, and particles; so to be in equilibrium, they must have the same temperature, pressure, and chemical potential. We're working at a fixed temperature by assumption, so that's not a problem. The pressures are the same too, since
$$
P = - \left( \frac{\partial F}{\partial V} \right)_{T,N}.
$$
By construction, the slope of the Helmholtz free energy is the same at points $A$ and $B$; thus, $P_A = P_B \equiv P$. Finally, we note that since these two points are connected by a straight line, we have
$$
\frac{F_B - F_A}{V_B - V_A} = - P,
$$
which can be rearranged to obtain $F_B + P V_B = F_A + P V_A$, or $G_A = G_B$, which implies $\mu_A = \mu_B$. So the two subsystems in the heterogeneous state are indeed in equilibrium.
As an aside: note that the condition $\left(\frac{\partial^2 F}{\partial V^2}\right)_{T,N} < 0$ is not satisfied by all the points between $A$ and $B$. If the Helmholtz free energy graph is concave down, this means that the system is unstable; but the converse is not necessarily true, since there are points along the graph for which $\left(\frac{\partial^2 F}{\partial V^2}\right)_{T,N} > 0$ but which do not represent states of minimal $F$ when compared to a heterogeneous state.
Best Answer
The more formal derivation of the van der Waals equation of state utilises the partition function. If we have an interaction $U(r_{ij})$ between particles $i$ and $j$, then we can expand in the Mayer function,
$$f_{ij}= e^{-\beta U(r_{ij})} -1$$
the partition function of the system, which for $N$ indistinguishable particles is given by,
$$\mathcal Z = \frac{1}{N! \lambda^{3N}} \int \prod_i d^3 r_i \left( 1 + \sum_{j>k}f_{jk} + \sum_{j>k,l>m} f_{jk}f_{lm} + \dots\right)$$
where $\lambda$ is a convenient constant, the de Broglie thermal wavelength and this expansion is simply obtained by the Taylor series of the exponential. The first term $\int \prod_i d^3 r_i$ simply gives $V^N$, and the first correction is simply the same sum each time, contributing,
$$V^{N-1}\int d^3 r \, f(r).$$
The free energy can be derived from the partition function, which allows us to approximate the pressure of the system as,
$$p = \frac{Nk_B T}{V} \left( 1-\frac{N}{2V} \int d^3r \, f(r) + \dots\right).$$
If we use the van der Waals interaction,
$$U(r) = \left\{\begin{matrix} \infty & r < r_0\\ -U_0 \left( \frac{r_0}{r}\right)^6 & r \geq r_0 \end{matrix}\right.$$
and evaluate the integral, we find,
$$\frac{pV}{Nk_B T} = 1 - \frac{N}{V} \left( \frac{a}{k_B T}-b\right)$$
where $a = \frac23 \pi r_0^3 U_0$ and $b = \frac23 \pi r_0^3$ which is directly related to the excluded volume $\Omega = 2b$.