What does 40VA rating for transformer mean?
Those 12V are the effective AC voltage. To get the peak voltage you need to multiply by the square root of 2, which results in about 17V.
40VA means that at an effective voltage of 12V it can handle an effective current of 40VA / 12V = 3.3A, which should be enough (though you neglected to tell us how much current the horn needs).
You need a rectifier and a smoothing capacitor in front of the linear regulator. The rectifier will reduce the voltage by about 1.4V, so at the input of the linear rectifier you will have a little more than 15V, with some ripple depending on the capacitor and the current drawn.
At 200mA and 10V difference between input and output your linear regulator will need to get rid of 2W of power, so it will need some cooling.
As for the horn, it depends on the specification of the horn, how precisely it needs those 12V. If it can tolerate some ripple you could use five 1N4004 diodes in series (5 * 0.7V = 3.5V, 15.5V - 3.5V = 12V) to lower the voltage. This is better than a resistor, because the voltage across a resistor depends much more on the load than for a diode.
40 VA is "Volts times Amps," which is typically close enough to 40 Watts for hobbyist purposes. The reason they say "VA" instead of "W" is that if your voltage lags (or leads) current, you get (VA) times (cosine of the angle between the V and the A), instead of just VA. If V and A are in phase (pure resistive load), then VA = Watts.
Transformers inherently handle only AC. Unless your tranformer is just plain broken (probably not), it is putting out somewhere around 12V AC as specified. If you're reading this with a DC voltmeter, it will read 0V since that is the DC average. Change the voltmeter to the AC Volts scale, and you should see somewhere around 12V, probably a little higher.
The AC voltage needs to be "rectified" to make it DC. A quick and dirty method uses a single diode and output capacitor. That basically passes the positive parts of the AC cycle and blocks the negqative. It will have some ripple, but will work since you're going to regulate it down to a much lower voltage. A better method is something called a "full wave bridge". This uses 4 diodes but harnesses both halves of the AC cycle to contribute to your DC power voltage. Basically, a full wave bridge takes the absolute value of the AC voltage.
As Starblue mentioned, either way the DC voltage will come mostly from the peaks of the AC voltage. These are sqrt(2) higher than the RMS voltage. The 12V spec will also be under maximum load, so the actual voltage will be a bit higher than 12V * sqrt(2).
2W is a lot of heat to get rid of in the regulator. That's too much for a TO-220 package in free air, and will require a heat sink. Alternatively, you could use a switching regulator if you feel comfortable with such things. It's more advanced, but is a worthwhile thing to learn.