Hi Robyn,
That oscillator circuit generates a "curved sawtooth" oscillation at the output, but the point in the circuit where the output is taken from is a "high impedance" point in the circuit. You can think of it as being "weak" or "flimsy". "Loading" it, by connecting resistors with values like 10k, will affect the circuit and cause it to stop oscillating.
This is why you kill the oscillator when you connect a voltage divider to it, and when you connect the ground of your sound card. A sound card has a typical input resistance of around 50k and this is too much load for the oscillator to drive.
So you do need to use a circuit called a buffer, which places almost no load on the oscillator circuit and provides an output that you can connect load resistors (such as a voltage divider with 10k and 1k resistors) to. That oscillator, as designed, is only intended to demonstrate the action of the oscillator itself; its output impedance is too high to be useful without a buffer of some kind.
Alternatively, you can use a voltage divider made from very high value resistors, which put very little load on the oscillator, but the final output will have an even higher impedance. It's not a good general approach.
The simplest type of buffer is called an "emitter follower". It's made from a single transistor; in this case an NPN such as a 2N3904 or BC547B, or any small-signal NPN transistor. A transistor with a high current gain, such as a BC547C/548C/549C, provides more buffering. Connect the transistor as follows.
Collector: connect to the positive supply rail.
Base: connect to the oscillator capacitor, i.e. the point currently labelled as the output.
Emitter: this is the output of the buffer. You need to connect it to the 0V rail through a resistor; you can use the voltage divider, as this provides the "emitter load" and also attenuates the voltage.
That's all you need to do. Just one transistor, and the two resistors that form the voltage divider.
I hope thiis is clear. If in doubt, look up "emitter follower" on Wikipedia.
Edit: BTW I'm not sure if you've realised this yet, but the "massive buzz" you hear is the actual oscillation of the circuit. A 36 Hz sawtooth sounds like a buzz. If you want the oscillator to run at a higher frequency, so it sounds like a tone, you need to reduce the value of the capacitor in the oscillator. The frequency is proportional to the reciprocal of the capacitor value, so for example, halving the capacitor value (from 100 nF to 47 nF) will double the frequency. A capacitor around 6.8 nF (also called 0.0068 uF) will give a frequency of around 500 Hz, which is a medium-pitched tone.
Edit 2: If you want an actual sawtooth waveform, rather than the curved waveform generated by this oscillator, you can replace the 100k resistor that charges the capacitor with a "constant current source" (see also Wikipedia). But there are better oscillator designs around; I wouldn't use this one except as an initial experiment.
Edit 3: That oscillator has a design error: it applies a large voltage across the PNP's base-emitter junction in the reverse direction, causing the junction to break down. (Any reverse base-emitter voltage above about 7V does this, because the transistor's base-emitter junction behaves like a zener diode.) This can damage the transistor. Using a power supply voltage of 6V will avoid this problem but you may need to change some resistor values to make it oscillate reliably.