Why is the voltage gain \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=-\frac{R_L}{R_E}\$
The minus sign in equation Vout/Vin = - R_load/R_emitter may appear counterintuitive. To understand its origin, remember that the network analysis here is performed for a linear circuit whereas the transistor circuitry is inherently non-linear. In order to be able to use Ohm's and Kirchhoff's laws, the circuit designer makes a number of idealizations, starting with an assumption that the voltages/currents in the circuit can be written as sums of DC voltages/currents plus (small additions of) AC voltages/currents. Or, alternatively, s/he examines small variations of voltages/currents near their "quiescent" values.

So, to fully understand the network analysis of transistor circuits, I would recommend you to learn from textbooks that use transistor models with current controlled current sources (CCCS)/voltage controlled current sources (VCCS) etc., the said models used to calculate biasing, gain, other parameters of small signal transistor amplifiers.

As for the common collector amplifier of the electronics-tutorials.ws tutorial, notice that for all practical purposes we can assume that C1 capacitance is sufficiently high, so that the AC component of V_base voltage is equal to Vin (Vin is an AC voltage source, I assume). In the Vin source's frequency region in which the C2 capacitance is sufficiently small to neglect current flowing thru a capacitor branch, an AC component of the current I_emitter is practically equal to an AC component of the V_emitter voltage divided by the R_emitter resistance. In the small signal approximation, V_base - V_emitter is constant (usually about 0.7V), therefore ΔV_base = ΔV_emitter, and ΔI_emitter = ΔV_base / R_emitter. Within a 1 percent error (beta = 100) I_collector = I_emitter; the AC components of emitter and collector currents are also equal (ΔI_collector = ΔI_emitter).

Vout is a voltage at the collector output of the transistor, as well as its AC component ΔV_load. Ohm's law gives us a voltage across the load resistor R_load measured relative to a reference point which is +Vcc pin of power supply (this means current flows out of R_load resistor); hence the minus sign in the equation

ΔV_load = - ΔI_collector * R_load. 

We arrive at equation ΔV_load/ΔV_base = - R_load/R_emitter.