My answer is mainly here to give some quantitative reasoning, but for the sake of completeness I'll also answer what my predecessors already answered.
Why do we divide by 4?
Imagine you see earth from the perspective of the sun. What you see is a disc. The area of this disc is $\pi r^2$ and therefore the total energy earth received from the sun is $S_0 \pi r^2 (1-\alpha)$, where the $(1-\alpha)$ takes into account that the solar radiation is partly reflected. As stated in the other answers, the idea is to distribute the energy over the whole surface of earth $4\pi r^2$. Thus, energy received per $m^2$ is $\frac{S_0 (1-\alpha)}{4}$.
Why is it reasonable to assume that the energy is distributed evenly over the whole globe at an instant?
The short answer is:
If we use this assumption we are usually dealing with timescales much longer than a full rotation of the earth. Thus, daily variations are negligible.
A quantitative example:
Let's compare the time it takes the atmosphere to adjust to an imbalance $N$ to the time it takes the earth to rotate once. A full rotation takes approximately $t_{rot} \approx 60\times60\times24s = 86400s$.
The imbalance problem requires a few explanations up front. Suppose we double $CO_2$ in the atmosphere. This causes a forcing of $F_{2 \times CO_2} \approx 3.7 \frac{W}{m^2}$ according to IPCC. We now want to know how long it takes the atmosphere to adjust to this forcing.
Let's setup a model: Following Gregory et. al. we model the imbalance linearly as
$N = F_{2 \times CO_2} + \lambda T$,
where $T$ is the temperature relative to some reference $T_0$ and $\lambda$ is the (negative) feedback parameter with units $\frac{W}{m^2 K}$. Thus an increase in temperature reduces the imbalance. Additionally we assume that the imbalance will lead to an increase in temperature over time (as done e.g. here)
$N = C\frac{\text{d}T}{\text{d}t}$,
where $t$ is time and $C$ is the heat capacity of the atmosphere. Combining the two equations above we find
$C\frac{\text{d}T}{\text{d}t} = F_{2 \times CO_2} + \lambda T$.
The solution to the differential equation above is
$T(t) = \frac{-F_{2\times CO_2}}{\lambda} \left(1 - e^{\frac{\lambda}{C}t} \right)$.
We can see the adjustment process takes an infinite time, however two thirds of the process are done within the $e$-folding time $\tau$ (when the term in the exponent is $-1$). The exponent is $-1$ if $t = \tau = \frac{-C}{\lambda}$.
Note that if $t$ is approaching infinity we have $T(\infty) = \frac{-F_{2\times CO_2}}{\lambda}$ (This is called Equilibrium climate sensitivity in the case of doubling $CO_2$).
What's left to do is estimating $C$ and $\lambda$. We can estimate the heat capacity of the atmosphere (just a column) to be
$C = c_p \frac{p_s}{g} = \frac{1005 \frac{J}{K kg} 10^5 Pa }{9.81 \frac{m}{s^2}} = 1.02 \times 10^7 \frac{J}{m^2 K}$.
IPCC tells us that $T(\infty)$ is likely to be between $1.5°C - 4.5°C$. Let's set $T(\infty) = 3°C$ and calculate
$\lambda = -\frac{3.7 \frac{W}{m^2}}{3 K} = -1.23 \frac{W}{m^2 K}$. Finally we find
$\tau = - \frac{1.02 \times 10^7 \frac{J}{m^2 K}}{-1.23 \frac{W}{m^2 K}} \approx 8.7 \times 10^6 s \approx 100\times t_{rot} $.
Caution
In reality the time scales differ a lot more. The oceans heat capacity is much higher than the atmospheres. However, the calculations above should convince you that even if all forcing changes atmosphere only time scales differ by at least $\mathcal{O}(2)$.
