The input range of your 5V regulator is 7 -25V. In case you want to use 7.2V as input voltage, there is only 0.2V margin. This is not OK, because if the input voltage drops let's say 500mV over the diode, to 6.7V, and you can be sure that the 5V output rail would be dropping also (at ~ 4.7V). This kind of unstable voltage on the supply rail can disturb the function of digital circuits and has to be avoided.
Also, you can improve the design with some small ceramic capacitors (e.g. 100nF) at each input, for your regulators. Place them close to the pin (a few mm). Have a look on the datasheet recommended schematic (page 3 in the 3V3 spec).
Other than that, your design seems OK. You can keep the 3V3 supplied directly from battery. It may heat up, but, it is specified up to 15V input. Just make sure you dissipate the heat properly. Make a little testing. It will not blow up from the first trial.
Depending on your input voltage, but at that low current, it makes no real difference. You would simply be changing which one gets the bigger energy wastage (in watts).
The LM1117 shown in your picture has a 5mA quiescent current, a minimum load current of 1.7mA, and typically up to 800mA regardless of the package. 15v Max input. You are using 9v, through a diode.
This is simple Ohm's Law calculations for Power (P = I * V)
Since they are Linear regulators, Current in is Current Out. For the first schematic, that means 9v - 0.7v (Diode Drop), 8.3v - 5v = 3.3v voltage drop at the 5v regulator. Then 5v - 3.3v = 1.7v voltage drop at the 3.3v regulator.
Then we can calculate heat dissipation. 1.7v * (30mA Load + 5mA Quiescent) = 0.06W for the 3.3v reg. For the 5v reg, we have to add the 3.3v's current load to the calculation. 3.3v * (8mA Load + 35mA 3.3v's Reg + 5mA Quiescent) = 0.159W. Grand total wasted power of 0.219W, with 75% of it on the 5v regulator.
For the second schematic, its almost the same. 3.3v drop for the 5v regulator, but 8.3v - 3.3v = 5v drop for the 3.3v regulator. Since the 5v regulator doesn't carry the 3.3v's load, it's just 3.3v * (8mA Load + 5mA Quiescent) = 0.043W. The 3.3v regulator though, 5v * (30mA Load + 5mA Quiescent) = 0.175W. Added together, it's 0.218W, with 75% on the 3.3v regulator.
In any case, the 0.218W in heat is nothing to be concern about. The regulators won't even get warm, let alone hot enough for a heatsink.
Best Answer
Here is the type over voltage protection circuit I would use if it was essential to protect against a greater than 3.3V input. This circuit uses a low cost voltage comparator to compare a 2.5V reference voltage against a divided copy of the input voltage and then will turn off a P-channel MOSFET when the input voltage goes above 3.3V.
You can change the reference voltage part to some other similar part at another voltage if needed and the scale the voltage divider resistors accordingly. One example of a low cost reference is a TLV431. A low voltage Zener diode could also be used but the precision of the circuit would be lost and the voltage divider would have to be tweaked to allow for greater margin above the 3.3V level.