If we assume continuity, infinite speed of sound, no viscosity, and laminar flow, then the key is in Bernoulli's equation. In general the pressure variation is very complicated, but to get some ideas how it works we can consider the pressure variation a very simple system that can be calculated analytically.
I came up with this simple system:
Let's say somewhere in the water far from the surfaces, suddenly a cavitation bubble of radius $R$ with pressure $P_0-\Delta P$ is produced. Where $P_0$ is the atmospheric pressure, let's assume that the scale of the system we are playing with is small enough that the pressure of undisturbed water anywhere is equal to the atmospheric pressure. Because of the pressure drop, water will start moving radially to fill the bubble as shown in the picture. If the flow is continue and laminar then we have the following flux relation
$v 4\pi r^2= constant$
$v=\frac{C}{r^2}$..........................................(1)
Now Bernoulli's equation gives
$P_0=P(r)+\frac{1}{2}\rho v^2=P(r)+\frac{1}{2}\rho \frac{C^2}{r^4}$
Because $v=0$ at a point far from the bubble. Boundary condition at $r=R$ gives
$P_0=(P_0-\Delta P)+\frac{1}{2}\rho \frac{C^2}{R^4}$
Eliminating $C$ and $\rho$ we get
$P(r)=P_0+\Delta P \frac{R^4}{r^4}$
As expected $r\rightarrow \infty$, $P(r)\rightarrow P_0$. Thus the pressure change fades away as we go farther from the source of disturbance.
In a more general case where the above assumptions still hold, the velocity profile of the flow equivalent to our eq.(1) is quite complicated. Eq. (1) can be replaced with a general flux equation which holds along a streamline
$vA=constant$..................................(2)
Where $A$ is the cross sectional area of a streamline portion. A typical flow's streamlines caused by a moving object look like this
We can view eq.(2) as an equation that holds in the moving object's frame. As we can see in the picture above, initially the streamlines are uniformly separated. Let's denote the cross sectional area of each slice of stream line portion as $A_0$, from here we know that if the cross sectional area of a stream line portion equals to $A_0$ then its pressure is unchanged or $P_0$. We can see that the streamlines near the object are denser, which means that their cross sectional area are smaller than $A_0$. Thus from eq.(2) we realize that the water is moving faster there, and Bernoulli's equation says that the pressure there is lower than $P_0$. As we move perpendicularly to the flow, away from the moving object the streamlines' cross sectional area get more and more similar to that of undisturbed ones and so does the pressure there. Therefore it can explain how the pressure disturbance decreases over distance from the source.
You must be aware that when two containers of water are connected by a pipe and if those containers are not airtight, water will flow between them until water-levels become equal. Obviously the container for feeding birds is open to atmosphere.
Now if the reservoir is not airtight (perhaps there is a hole or cut somewhere, through which air can flow in directly from ambient to reservoir) then the necessity for equalization of water levels explains the results of experiments #2, #3, and #4. It can also explain results of experiment #1 if water in the reservoir wasn't initially deep enough to cause overflow from the container, when water-levels equalized.
But say you have been extremely careful to ensure that there is no hole or cut on the reservoir that can allow air in directly from ambient to reservoir. In this case what you have observed can still happen, and may be qualitatively explained thus: As water is emptied out of the airtight reservoir the air initially inside the reservoir (atop the water surface) expands resulting in a lower pressure there. Therefore the body of water inside the reservoir experiences two counter-forces: one is due to the hydrostatic head caused by the difference in water levels inside the reservoir and the container and which causes the water to flow out of the reservoir; and second is due to the difference in ambient pressure outside and low pressure zone inside the reservoir atop the water surface that prevents the water from flowing out. At some level difference between the water in the reservoir and the container (and for a given geometry of the reservoir), the two forces become equal and flow shall stop. However by the time this point of no-flow is reached the container might already be overflowing.
This idea may be formulated precisely. Let $V_0$ be the initial volume of air inside the reservoir atop the water surface, which is also at atmospheric pressure. Let us measure all heights from the ground on which the reservoir and the container rest. Let us take the container to be sufficiently deep that water cannot overflow. When water flows between them and equilibrium is reached, let $h_R$ and $h_C$ be the water levels in the reservoir and container respectively, measured with respect to the ground. Corresponding to final water level in the reservoir $h_R$ the volume of air atop the water surface in the reservoir shall be $V$, so that air in the reservoir has expanded from $V_0$ to $V$. If the temperature change is insignificant so that we may assume it to be constant, using ideal gas law for air (which is a good approximation) we have magnitude of pressure difference w.r.t. ambient as:
\begin{align}
|\Delta p|=mRT\left( \frac{1}{V_0}-\frac{1}{V} \right)
\end{align}
in which $m,R,T,$ are respectively mass, gas constant, and temperature of air inside the reservoir. In equilibrium this pressure difference is counterbalanced by the hydrostatic head $\rho_{water}g(h_R-h_C)$. This gives:
\begin{align}
|\Delta p|=\rho_{water}g(h_R-h_C) & =mRT\left( \frac{1}{V_0}-\frac{1}{V} \right)\\
\therefore\quad h_C & =h_R-\frac{mRT}{\rho_{water}g}\left( \frac{1}{V_0}-\frac{1}{V} \right)
\end{align}
If the actual height of the container is less than $h_C$ then the water will overflow. Remember that $V$ in the expression above is a function of $h_R$ and each configuration in your experiments shall require a separate calculation.
P.S. I am not sure why you do not wish to employ valves. Float valves are not very expensive and do not require power input to work. There is another (not very pleasing to me) alternative to maintaining a nearly constant water level without using any extraneous devices. Take the container with as small a floor area as is feasible, and choose a reservoir with as large a floor area as is feasible. Then for a given amount of water flowing from reservoir to container the level change in the reservoir shall be smaller the larger the ratio of area of reservoir to container. To be precise, let $A_R,A_C$ be the area of reservoir and container respectively. Let $\dot{q}''$ be average rate at which volume of water is consumed from the container per unit area per unit time, due to birds drinking from it and due to evaporation. Let us assume that the rate at which the container loses water is small enough (which seems a reasonable assumption to me) so that water levels in reservoir and container equalize practically instantaneously, so that at all times water levels in reservoir and container are equal. Then $\dot{q}''A_C\delta t$ is the volume of water lost from reservoir in small time duration $\delta t$. This causes a dip in water level $\delta h$ in reservoir and container. Mass balance then gives: $(A_R+A_C)\delta h=\dot{q}''A_C\delta t$, therefore the rate at which height of water in container changes is: $\dot{h}=\dot{q}''/(1+A_R/A_C)$.
Best Answer
What you're missing is some practical aspects of modern plumbing. Every modern plumbing fixture(toilet, shower/sink drain) has a trap or "u-bend" in the piping before it connects to the sewer branch or main pipe. This allows a pocket of water to sit in the bend of the pipe to keep sewer gases from escaping back into the room when the fixture is not being used. This water pocket obeys the equilibrium principle you know. When the fixture is used, the water above disturbs this balance and everything flows into the sewer pipe.
During your sewer back-up event, something has happened downstream to provide enough pressure reverse the normal flow, and the material will seek any free outlet (e.g. your shower drain). These events are usually very short-lived, so probably provided enough pressure to push stuff through the u-bend in the toilet and shower, then ceased. Once that pressure was gone, the normal physics applies and the liquid levels then reach equilibrium with their respective u-bends. Your shower basin, despite being lower than the toilet, contained the spill because it had a larger area to contain the volume of the spill.
Source: Plumbing design engineer.
More about plumbing traps here.