The answer to your precise question is no: it is not always the case that $F = N_2\cap [R,R]$. A nice way to see this is by fixing a field $k$, and constructing the universal example of a $k$-algebra equipped with a square-zero commutator. That is, $R = k\langle a,b\rangle/((ab - ba)^2)$.

In that $k$-algebra, the element $ab - ba$ is certainly a square-zero commutator. The question is whether that element is equal to $yx$ for some pair of elements $x,y\in R$ such that $xy=0$. You can rule the existence of such a pair $x,y\in R$ by grading $R$ so that $a,b$ are each in degree $1$. Then the relation $(ab - ba)^2$ in $R$ is homogeneous of degree $4$. So if you have $yx = ab- ba$, then:

$x(ab - ba) = xyx = 0$, so $x$ must have no nonzero terms of degree $<2$, and

$(ab - ba)y = yxy = 0$, so $y$ must have no nonzero terms of degree $<2$.

But $ab-ba$ is homogeneous of degree $2$, so there's no way to multiply $x$ and $y$, each with no terms of degree $<2$, to yield $ab-ba$.

As an aside, in the case where $k = \mathbb{F}_2$, a certain $k$-algebra quotient of $k\langle a,b\rangle/((ab - ba)^2)$ arises quite naturally in topology:
the $k$-algebra
$$ k\langle a,b\rangle/((ab - ba)^2,a^2, aba - b^2)$$
is isomorphic to the subalgebra of the Steenrod algebra generated by the Steenrod squares $Sq^1$ and $Sq^2$. The element $ab - ba = Sq^1 Sq^2 - Sq^2 Sq^1$ is one of the famous Milnor primitives, usually denoted $Q_1$. That element is also an example of a square-zero commutator which is not in your set $F$.

Thanks for asking an interesting question, which I enjoyed thinking about.