Linear behavior from 0-10V and steady at 3.3V from 10-24V
I've come up with 3 alternatives that will most likely suffice. As someone who has "very basic electronics background", I don't assume that you've got diodes or operational amplifiers laying around. So that's why I'll propose several circuits. 



First off is the circuit that will follow your graph and overshoot by 1 mV. 



There's two 10k in parallel instead of one 5k because if you use three resistors of the same kind then it's very likely that you'll reach 3.33 V if you feed it 10 V. I chose 10 k because it's not too small or too high. As we say in Swedish, it's "lagom". 

If you look at the bottom graph in the image above you will see the same graph that you were asking for. 

The operational amplifier (op-amp), is used as a comparator and clipper, if your input is below what you put into the minus sign of the op-amp (3.3V) then the op-amp won't do anything, if you go above 3.3 V then it will activate the N-mosfet and pull down your output so it stays at 3.3 V. This answers your question, however you might not have an op-amp or an N-mos or NPN BJT. 



Secondly comes the circuit that is maybe "good enough" for you. 



It's not exactly what @inkyvoyd proposed because I used several diodes instead of one zener because not everybody keep a 3.3 V Zener at their house. Diodes ([0.5 - 0.8] V voltage drop) however may lay around, or a blue LED, they usually have around 3.4 V voltage drop.

Again I put two 10 k resistors in parallel for ease of use. 

As you can see at the bottom of the image there's a voltage graph, the bottom one shows the output, and it's not as sharp as the first graph in the first image. So that's what you might expect with 10 k ohm resistors, if you would use three 10 ohm resistors then it might be near the same sharpness as the first image, but 10 ohm resistors will burn up when you feed it 24 V. So don't do that. 

If you however would replace the diodes with one 3.3 V zener diode then it would behave nearly identical like the op-amp solution. 



Thirdly, here's the schematic that I want you to use instead. 



instead of making a non-linear curve that you want, why not make a linear curve and make up for it in software? 

So 0 V input gives 0 V output, 24 V input gives 3.3 V output. This will result in 10 V input giving 1.375 V output. Then you just scale that in software. 

It's by far the easiest solution of the three proposed solutions, just two components. However it requires a 1.6 k ohm resistor which is most certainly not something a typical person got laying around.

So perhaps a 10 k potentiometer (or higher, 100 k would be nice) could be used instead, and trim it so it gives 3.3 V if you feed it 24 V and you're set. One component that you most likely have laying around. DO NOT use a 1 k ohm potentiometer. 

If you use a 1 k ohm potentiometer then it will burn off \$\frac{24^2}{1000}=0.576W\$. 0.5 W for a sensor is nasty! Also it might heat up to nasty high temperatures. 



Here's the schematics in simulation form so you can see for yourself and change values if you want.