Why does this capacitor shorten the clock pulse?
To understand intuitively the operation of this CR differentiating circuit, think of the capacitor as of a kind of "rechargeable battery". So two voltage sources are connected in series... and this "battery" of two voltage sources is loaded by the resistor R.

1. Initial zero input voltage. Assume that initially there was no input voltage long enough and the capacitor is fully discharged. So there is no voltage applied across the resistor... and there is no current flowing through it...

2. Positive input voltage. The input voltage rises sharply but the voltage across the capacitor cannot change immediately... and is zero at the first moment. The input voltage source faces no obstacle; the whole input voltage is applied through the capacitor to the resistor and appears at the output (forming the positive "tip of the "needle").



The capacitor quickly charges... its voltage increases thus increasingly hampering the input source... and the voltage across the resistor (the output voltage) decreases. It is an interesting situation - the input source supplies the capacitor "source"... but the latter opposes it... and finally completely neutralizes its voltage (as in the proverb, "Feed a dog to bark at you":) The "battery" consists of two equivalent but opposing voltage sources. There is no current flowing... there is no output voltage.

3. Zero input voltage. The input voltage drops sharply to zero (short connection, not open circuit!). The capacitor "anti-voltage" cannot change immediately... and is equal to the previous input voltage. There is only one acting voltage source - the capacitor... its voltage is applied  through the input source (short connection) to the resistor and appears at the output (forming the negative "tip of the "needle"). The current flows in an opposite direction... and the output voltage is negative...



The capacitor quickly discharges... its voltage decreases up to zero. There is no current flowing... there is no output voltage.