You will need to calculate the required track width for the parameters given. KiCad’s Calculator Tool (or Saturn) will be helpful.
Just a small comment on terminology to avoid misinterpretation and confusion :
In a Schematic, you wire symbols.
In a PCB, you place tracks between footprints.
Components are in data sheets or on a finished board.
Memorize this terminology and you are well on your way. 
Hello Jonas1893 and welcome to the fun that is the creation of a PCB.
My method.
I make a folder in my PC of all of the manufacturers components specifications.
I find the PHYSICAL pin out diagrams for all parts and print them out. For some I pencil in additional pin descriptions if needed (such as make a gpio pin the Activity LED pin).
You must do this for the connectors too.
I also consider what kind of enclosure there is going to be and where the connectors will be.
If you can put all connectors on one edge you may make assembly easier and the enclosure design easier.
On paper I print out the schematic parts making them large enought to be easy to read. T
Place the connectors on the edge of the schematic or around the schematic preserving the order of the same on the enclosure. As much as you can, make this similar to the placement you expect on the enclosure. If you do this you will begin to “see” where next the electrical components get placed and oriented to minimize lines crossing on the schematic which will help when it comes to the PCB.
You can see that I am trying to gain insight into the PCB design as early as I can at the schematic design level.
For ICs I draw the capacitors on every power pin to ground next to the power pins. This makes sure that the all important design of power supply “bypassing” is not forgotten. I decouple the power through a series resistor EVERY SINGLE TIME. Suppose I have a 555 Timer. I would have a power capacitor to ground of for example 1nF and a 22uF and I would label the net 5VTIMER. Then from that net a series 1 Ohm resistor that connects to the raw +5 net.
Place the other components around the 555 which set the behavior (Make it an oscillator or one shot or ???) and place some descriptive text for the function.
You will find a lot of schematic symbols with input pins on the left and output pins on the right. I advise against this method of schematic sysmbols. Test my idea yourself. Do a simple 555 timer circuit from scratch with schematic symbols that look like the real part and again with the symbols with input pins on the left and output pins on the right.
This method helps me. I believe it also helps other persons who may work with your PCB (such as factory test or troubleshooting) because they will more quickly come up to speed on component location
Always put a power on indicator LED in the design.
Always put a activity indicator LED in any design with a micro controller.
Always put labeled power supply test points in every design.
I always feel I get best results with this design approach.
Your mileage may vary.
More consideratsion as you become experienced
For real products you must consider safety. Safety means that no component failure can cause a shock or fire hazard. Imagine that every single capacitor (one at a time) might become leaky and start conducting DC current with what ever value causes the worst thing to happen. No part must become a shock or fire hazard. This generally means you need current limited power supplies. The current limit may be as simple as a fuse that destroys the power supply such as is common in a wall wart. But some where in every design is a current limit. Make sure you know what it is and that every single component failure (short, or open or any value in between) does not cause a fire or shock hazard.
The resistor “decouples” the power supplies from the distribution at high frequencies. The “bypass” capacitors provide the path for current for the high frequencies local to every load.
You could think of a fast microcontrollers as a current load that is switching spikes of current at the clock rate. That current must not come from a node shared with all the other signals or they will be modulated by the delta voltage that current load causes.
The power distribution risks becoming a giant intermodulation of all signals (especially the edges so to speak).
If you put a capacitor on every single power pin with a very small geometry (low inductance) the current will so to speak choose to be sourced by the capacitor rather than the giant mixing node that is the power distribution. The resistor helps. With out it you have to imagine the equivalate circuit where the coper traces are now the resistance and you have a much higher Q distributed mess of reflected and mixed current signals.
Again the resistor helps. In general you want the largest resistance compatible with normal operation.
If for example your 555 timer is driving on the output a 1K load then you can get away with even higher than one Ohm. If you are power a motor drawing an amp, or one hundred, then you will need lower impedance and may need to resort to ferrite beads which are more expensive. Bottom line you have to engineer your power distirbution. Just like the PCB it is one aspect of the design for which you will have to understand YOUR system as you put it together. No one else can know what your system is doing.
Hi, thanks for the advice. Now everything works I think
Thank you! I will memorize it for sure 
Thank you so much for your detailed explanation!
I think I finished now with the PCB. Can someone have a quick look on it if it seems ok or not? That would be awesome
I also attached one picture. In green I marked the IRF7343PBF MOSFETs. I’m not sure whether I connected them correct or not. Did I place the tracks correctly (see in green)
Best,
Jonas
BLDC_Ansteuerung.kicad_pcb (334.5 KB)
BLDC_Ansteuerung.kicad_prl (1.1 KB)
BLDC_Ansteuerung.kicad_pro (9.1 KB)
BLDC_Ansteuerung.kicad_sch (102.8 KB)
fp-info-cache (1.6 KB)
Start by not overlapping texts in your schematic. Things like below should be fixed before you post your work:
I doubt the circuit below will work.
It’s quite likely that both Q6A and Q6B will be open at the same time during the switching for a short period.
It’s nice and all to experiment a bit with this in simulations or on a breadboard (make small breakout boards for the SMT FET’s) but for a “real” circuit it’s much better to use a MOSfet Gate driver chip that replaces your 8 small fet’s, and also have built in circuitry to prevent cross-conduction of the big IRF FET’s.
I have not analyzed it properly, but your bootstrapping with C1 and C3 is probably not going to work either. It may lift the cathode of D1 to a voltage close to 24V and that is dangerously close to the max gate voltage for Q5A and Q5B.
Have a look at MOSfet Gate driver chips such as the IR2101. There are many similar IC’s on the market. DRV8303 is another one for 3 half-bridges.
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Hi @Jonas1893
I don’t know if it is a project requirement to use the layers for the text, but your text for the footprints is unclear.
For display purposes, I’d hide ${REFERENCE} … see red arrow and magenta box and I’d change the layer for value … green arrow (double left click, scroll to F.Silk and click).
This will make the PCB much more easily read.
Perhaps then adjust the positions of Ref & Value.
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I would like to restate what paulvdh wrote regarding Q6A and Q6B.
Bad things could happen if every Q6A and Q6B were to both be on and conducting at the same time.
TO get at this we ask, what would be the current through Q6B if they were both on?
If we pretent the on resitance of the MOSFETs is zero and the trace resistance is zero then the current load on the 12V supply will be 12/0 or infinite.
Do you see the problem?
Another fun thing is to look up the on resistance of the FETs.
We see ranging between 0.043 to 0.170 for the parts.
So the current is going to be something like about 12V/(0.2) or still an absurdly large 60 Amps.
Another fun thing is to estimate the copper trace resistance. A useful thing to know is to analyze the resistance of a trace using an estimate of it’s geometry and the idea behind Ohms/Square.
For a board fabricated with 1 oz copper the Ohms/Square is about 0.5mOhm.
If your trace is 1mm x 1mm the resistance is 0.5mOhm. If your trace is 5cm x 5 cm it is the same, 0.5mOhm. If your trace is 1mm x 40mm its resistance is 40X0.5= 20mOhm.
So of both transistors turn on the current through the transistors and the 20mOhm trace will be 12/(0.2 + 0.02) or 54.54 Amps. Still an absurdly large current. Such a current will distroy the parts quickly. If the transistors fail short circuit (which is a common failure mode) then only the trace resistance limits the current and it will go to 12/ 0.02 = 600 Amps. If 600 Amps went through a trace 1mm wide it would nearly instantly melt and as we say “let out the magic smoke on which all electronics work.”
This is why fuses are put in power supply circuits so that if the unexpected happens current is limited (by the fuse) to a safe (for the PCB traces) value.
Hope this was useful.
P.S. an article on estimating PCB trace resistance: https://www.edn.com/counting-squares-a-method-to-quickly-estimate-pwb-trace-resistance/
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