Ok, now I'm understanding.
Yeah, all you need is a switch debounce.
When you press a switch or flip a switch, although you may not realize it, the switch actually momentarily opens and closes, or closes and opens, several times before coming to its final state. You need to have a circuit that takes the initial change and ignores the rest. That's a switch debouncer.
Currently you have an R, a C, and a Schmitt trigger inverter. Assuming your counter is a CMOS device, it likely has a clock input that responds to the rising (leading) edge of the clock signal. Hence, you'll need your clock to go from low to high when you push your switch/button. Therefore, you'll need your input to your inverter to be normally high and then go low with the button.
For a normally high input, you'll want to pull up the input through your R to Vdd (your supply). Then you'll want your C to discharge quickly through your switch/button to Vss (Ground) and then charge again through your pull up R back to a high upon release. The charge time is determined by τ=RC.
Hence, at power up, your input will go high keeping your output low, and it willl charge your C according to τ. Then, when you press your normally open switch, which is across C, the C's charge gets quickly shorted to Vss and pulls your input low which, in turn, sends your output abruptly high and triggering your counter, via the leading edge of the output pulse (your rectangle). BUT...
As your C was charged, for that brief instant when the switch is closed, your input isn't going to go to Vss, 0v, but, rather, -Vdd, for it will look like a battery that's been reversed. This may cause a problem. (I'm not sure what a Schmitt gate input looks like, but good practice is not to use the input clip diodes onboard.) To limit the negative spike, you should add a diode from the input to Vss, with anode to Vss. This way the -Vdd spike will cause the diode to conduct (anode is at Vss=0v), bypassing the input of the inverter with the low impedance of an on diode.
However, you may have a normally closed switch. (You didn't say.) If this is the case, then you still need that normally high input (as an inverter is stilll an inverter), but you'll need the opening of your switch to pull the input low. This seems to me to be a little more tricky. I'll have to think on it, if that's what you need.