One should imagine solving this problem on a planar graph.
The force $\vec{F}$ is strictly a horizontal force (as indicated by the word problem), so one could write this as
$\vec{F}= \langle |\vec{F}|,0\rangle$. So $\vec{F}$ does not exert a vertical force onto the floor.
The force of static friction is said to be pointing in the negative $x$-direction. So $-\vec{f_s}=\langle -|\vec{f_s}|,0\rangle$.
The normal force $\vec{F_N}$ is the vertical force exerted by the floor. So $\vec{F_N}=\langle 0,|\vec{F_N}|\rangle$.
Finally, you have the gravitational force $m(-\vec{g})$ acting on the object, i.e., it is $\langle 0, -m |\vec{g}|\rangle$ (where without loss of generality, one could think of $\vec{g}$ as $\vec{g}=\langle 0, 9.8 \mbox{ m/s}^2\rangle$). The minus sign appears since the gravity is pointing downwards.
If the object were stable (non-moving), then we add these four vectors to get the resultant force vector $\langle 0,0\rangle$.
That is,
$$
\vec{F} - \vec{f_s} + \vec{F_N} - m\vec{g} =
\langle |\vec{F}|,0\rangle + \langle -|\vec{f_s}|,0\rangle + \langle 0,|\vec{F_N}|\rangle + \langle 0, -m |\vec{g}|\rangle =\langle 0,0\rangle.
$$
Since we add vectors component-wise, we see that
$$
|\vec{F}| -|\vec{f_s}|+0+0 = 0 \mbox{ and }
0+0+ |\vec{F_N}| -m |\vec{g}|=0.
$$
So
$$
|\vec{F}| -|\vec{f_s}| = 0 \mbox{ and }
|\vec{F_N}| -m |\vec{g}|=0.
$$
Since the $y$-components of $\vec{F}$ and $\vec{f_s}$ add to zero, we have $\vec{F} - \vec{f_s} = \langle 0,0\rangle$, and since the $x$-components of $ \vec{F_N}$ and $m\vec{g}$ add to zero, we have
$\vec{F_N} - m\vec{g} = \langle 0,0\rangle$.
Thus, the two equations
\begin{align*}
\vec{F} - \vec{f_s} &= \langle 0,0\rangle \\
\vec{F_N} - m\vec{g} &= \langle 0,0\rangle \\
\end{align*}
are precisely the two equations derived in the png attachment.
Now, notice that static friction is denoted by $-\vec{f}_{s,\max}$, and in the case when the object is moving, i.e., it has a nonzero displacement vector (in the $x$-direction), then that is when we have the two equations:
\begin{align*}
\vec{F} - \vec{f}_{s,\max} &= m\vec{a}= \langle m|\vec{a}|,0 \rangle, \\
\vec{F_N} - m\vec{g} &= \langle 0,0\rangle, \\
\end{align*}
where $\vec{a}$ is the resultant acceleration of the object in the positive $x$-direction.
In the case when $-\vec{f}_{s,\max}$ is the $\textbf{maximum}$ static friction $-\vec{f}_{s,\max}=\langle -\mu_s m|\vec{g}| ,0\rangle$, then this would mean that the object is not moving, i.e., the acceleration vector $\vec{a}$ is the zero vector $\langle 0,0\rangle$.
