Let me try and explain as simply as I can, and hope I don’t confuse you more!
So how by just sweeping the wings back, decrease the velocity of the air at the peak of the camber line?
It doesn’t, the speed of the air passing over the peak of the camber line is (broadly) unchanged, but wing sweep changes the angle at which the airflow hits the leading edge of the wing.
I don’t know how good you are at Maths, but we can break down (resolve) the motion of the air across the wing into two components. One component that flows directly across the wing at right angles to the leading edge (chordwise flow), and the other component, which is at right angles to the first component, that flows along the wing from root to tip (spanwise flow).
So on a swept wing, the speed of the air passing over the peak of the camber line at right angles to the leading edge of the wing has been decreased.
Very broadly, it is only the component of velocity across the chord of the wing, (chordwise flow), that is causing the shock wave, and interests us in terms of critical mach number.
In your example of an aircraft with wings swept at 30°, if the aircraft is travelling at V knots, the speed of the air flowing across the chord of the wing has been reduced to V*Cos 30° knots, or around 87% of the aircraft’s forward speed.
If that’s a little difficult to understand, try this. Pick up a ruler and pretend that it’s a wing. Mark a point at the middle of one of the (long) edges, and draw a thick black line backwards from that point, at right angles to the long edge, across the ruler to the other edge.
Hold the ruler between finger and thumb at one end, pretend it is a wing on an imaginary aircraft, and try moving it to different angles of sweep. Can you see that the speed of the airflow over this line will vary as the angle of sweep varies?
The speed of the air flowing along this line will be the same as the aircraft speed when the wing is fully forward (sweep = 0°) and will be zero when the wing is fully aft (sweep = 90°). In between, if the angle of sweep is x° , the chordwise speed will be V*Cos x°.
If we sweep the wings to 30°, whilst still flying the aircraft at Mach 0.75, the spanwise flow of the air will reduce in speed to Mach 0.75 * Cos 30°, roughly Mach 0.65.
Conversely, we could keep the chordwise flow at Mach 0.75, but increase the speed of the aircraft, from Mach 0.75, to (Mach 0.75 / Cos x°).
In your example, the Critical Mach number of the aircraft with an unswept wing was Mach 0.75, which produced a chordwise flow at Mach 1.0 at the peak of the camber point. Sweep the wings by 30°, and you can increase the aircraft speed to (Mach 0.75 / Cos 30°) which comes out to Mach 0.86.
I hope this helps,