Kirchoff's Voltage law states that the sum of the pd drops at each component in a closed loop is equal to the emf supplied by the battery. If I'm imagining it correctly, the electrical energy flows from the battery around the circuit (the electrons don't move around the circuit at near light speed to deposit this energy; they only move at rates of centimetres per second). Now consider a simple circuit with one resistor and one with two resistors. When the energy goes around the first circuit, it is all lost to the single resistor. However, in the second circuit, shouldn't there be the same outcome? If not, doesn't that suggest that the battery somehow 'knows' a second resistor is there and decides to only supply half of the emf at one resistor and half at the other? Or am I totally misunderstanding something?
If I understand your view
If I'm imagining it correctly, the electrical energy flows from the battery around the circuit
energy should flow from one pole to the other, leaving some portion behind along the way. From which pole to which?
Actually energy (not current) flows in a different way. It exits from the battery and flows to resistor(s), but not within wires. It flows in empty space between battery and resistors, guided - to say so - by the wires. Energy flow is given by Poynting vector
$$\vec S = \vec E \times \vec H.$$
I understand it's hard to visualize, without practice and with no help by figures.
Before you connect (close) the circuit, the electric fiels inside the wire and resistors are zero. Once you close it it becomes different than zero, and that 1) provides a path for the electrons to enter the wire and 2) allows the electrons to move at some speed given by the electric field. The resistances in the circuit are equivalent to a change in the electric field. If you have only one resistance the electric field will change one way, and if you have two it will change still more (it will make it weaker), slowing down the electrons and reducing the current.
First, in your parenthetical comment you state “the electrons don't move around the circuit at near light speed to deposit this energy; they only move at rates of centimetres per second”. Your comments about the speed are roughly correct, and further they imply that it is not the electrons that deposit the energy at all. In fact, the energy is transported by and deposited by the fields, not the electrons.
This concept is important for a couple of reasons. First, it explains why flipping a light switch results in the light bulb turning on almost immediately, and second it explains your question about how the various parts of the circuit appear to “know” about the circuit configuration. The information about the changes circuit configuration is carried in the fast electromagnetic fields.
In the case of adding a resistor in series. If the same current continues to flow momentarily then there will be a large voltage drop across the new resistor. The combined voltage drop will be communicated to the battery through the fields. The voltage source will respond by dropping the current. This reduced current will then be communicated back to the resistors which will drop their voltage according to Ohms law. This continues very rapidly through changes in the field until the voltage at the source is correct and follows KVL and the current through the resistors follows KCL and Ohm’s law.
One of the assumptions of circuit theory is that the circuit is so small compared to the speed of light that all of this occurs instantaneously. This makes it appear that components have mystical knowledge of the whole circuit. But in fact, this simplification is merely hiding the details of the information being communicated quickly through the fields