Rule 1 isn't a "good idea", it isn't a "guideline", it is a fundamental tenet of transistor physics. If for any reason (during normal usage) it is unable to hold then the circuit will not operate.
As for the lamp, it is a purely resistive element. It should have 10V across it, but thanks to the transistor it won't. So the transistor gets 0.2V and the lamp gets 9.8V and reality is saved.
The BJT collector current equation is
$$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CB}}{V_A}\right)$$
where \$V_A\$ is the Early voltage. But, this formula is often written as
$$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CE}}{V_A}\right)$$
Thus
$$\frac{\partial i_C}{\partial v_{CE}} = \frac{I_S\cdot e^{\frac{V_{BE}}{V_T}}}{V_A} = \frac{i_C}{V_A + v_{CE}}$$
This is clearly a non-linear function of the collector-emitter voltage and collector current so this cannot be interpreted as a conductance.
However, for small changes around some fixed value of collector current \$I_C\$ and collector-emitter voltage \$V_{CE}\$, we can write
$$I_C + i_c \approx I_C\left(1 + \frac{v_{ce}}{V_A + V_{CE}} \right) = I_C + \frac{v_{ce}}{r_o}$$
where
$$r_o = \frac{V_A + V_{CE}}{I_C}$$
We call \$r_o\$ the collector-emitter dynamic, or differential or small-signal resistance.
It is not a true resistance since it is not constant but, instead, varies with the operating point of the transistor as can be seen by the formula.
Best Answer
The most important difference is that for the common collector (that's the one with the load on the emitter side) you'll need a higher drive voltage. While for the common emitter 0.7 V is already enough, for the common collector the voltage must be 0.7 V + the voltage across the load.
Suppose your load is a 12 V relay, and you also supply 12 V to the collector. If you want to control that by a 5 V microcontroller then that 5 V is the maximum you can supply to the base. The emitter will be 0.7 V lower, that's 4.3 V, which is too low to activate the relay. The voltage can't go higher, because then there wouldn't be no base current anymore. So if the load voltage is higher than the control voltage you can't use common collector.
Also different is how you calculate base current. Suppose you apply 5 V on the base, the load on the emitter's side is 100 Ω and the transistor's \$h_{FE}\$ is 150. Maybe you would expect the current to be 4.3 V/100 Ω = 43 mA. That won't be the case. A base current of \$I_B\$ will cause 150 \$\times\$ \$I_B\$ through the 100 Ω resistor, not \$I_B\$. Therefore the created voltage \$V_E\$ = 150 \$\times\$ \$I_B\$ \$\times\$ 100 Ω. So the resistance seen by the base current is \$ {R_E}^{'} = \frac{V_E}{I_B} = \frac{150 \times I_B \times 100 \Omega}{I_B} = 150 \times 100 \Omega = 15 k\Omega\$.
So that 100 Ω resistor will cause a base current of only \$\frac{5V -0.7V}{15 k\Omega}\$ = 290 µA.
That's why you often won't need a base resistor in common collector configuration. You will need one though if the load consists of LEDs for instance, because contrary to the resistor these will cause a more or less constant voltage drop.