I think I am looking for a specific DC relay switch

Just to add to my comment with the circuit diagram.

The first circuit shows the DRL(day) connected to the common pole of relay 2. (The way I think you have wired your circuit) With relay 1 unenergized the ACC(+) has a path through relay 1 and relay 2 switches to DRL(day). However, when relay 2 is energized all that happens is DRL(day) and DRL(night) are connected together. There is no path for the current into either.

The second circuit shows what happens if you swap the connections of the COM and NC poles of relay 2. Now current has a path through the switch contacts and will either switch to DRL (day) or DRL (night). Note also I have added (snubber) diodes across the coils of the relay (e.g. 1N4001 types). These prevent spikes of back emf when the relay coils are turned off. Normally they are put in to prevent damage to driver transistors but they also function to prevent arcing across mechanical switches which would lead to premature failure.

As for the second (extra) bit of the question I wouldn't use a resistor as it would waste a lot of power. I would use a suitable P channel MOSFET with PWM (pulse width modulation) to control the power available to the bulb.

Additional edit (A PWM circuit)

The circuit above is a simple PWM using a 555 timer (CMOS type)and a power P channel MOSFET (transistor). There are many suitable devices such as the FQP27P06 which is rated for 27A, 60V. The main thing to look for is high current capability (>20A), low Rds (Drain-Source resistance on turn ON - typically < 0.1R - the lower the better) and a rated voltage >40V.

The circuit works by switching the MOSFET ON for a short period (mark) and then turning it OFF (space). The 555 astable oscillator repeats this very quickly (frequency controled by C2) so you don't see any flicker in the bulb. Setting the 100k variable (preset) to mid point should give half power (equal mark/space).

Because the MOSFET is turned either ON and OFF their is very little power dissipated even though it controls a large current. For a device with Rds = 0.01R and a current of 10A this waste power (heat) will only be 1W. (= I^2R) for the time it is ON. There is no power dissipated when it is OFF so the average power lost over the cycle will be less than 1W (runs nice and cool).

The battery that is currently switched in must keep the other battery switched out until the first battery is drained, and once switched over must stay that way until the other battery drains (even after the first battery is replaced). This implies a symmetrical bistable circuit.

Here is a possible solution using relays:-

^{simulate this circuit – Schematic created using CircuitLab}

Each battery has a relay with a normally closed contact that keeps the other battery (and its relay) switched off. Whichever battery is connected first will have priority - until it is drained, then its relay will turn off and give control to the other battery.

This circuit has a few potential problems.

1. When the drained battery's relay releases and connects the other battery the voltage will rise, which may cause the relay to pick up again before the other relay has time to operate. If this happens the relays will chatter instead of switching over cleanly.

2. The relays may become sensitive to mechanical shock or vibration when close to the release point, and the cutoff voltages may not be well defined.

3. Relay contacts increase resistance and get noisier with use. Eventually the circuit may become unreliable.

For these reasons I would consider using a solid state circuit (comparators and logic gates or an MCU) to monitor the battery voltages and control the switchover, and MOSFETs to switch between batteries.