How to adjust output current on a buck converter?
You don't "adjust about 3 A output current".  The job of the regulator between whatever power voltage you have and this Beagle thing is to provide a steady 5 V.  How much current the load (the Beagle thing in this case) draws is up to it.

According to your specs, it can draw as much as 3 A.  In that case, your 5 V supply has to be able to supply 3 A, but that doesn't necessarily mean it will always be supplying 3 A.  That is probably a peak spec, and part of the time the load won't draw that much.  That's fine though.  Your supply will keep its output at 5 V as long as the load draws 3 A or less.

What you need, therefore, is a buck switcher that can take around 9 V in, make 5 V out, and can supply up to 3 A at that 5 V.

Note that you are asking for 15 W.  That has to ultimately come from the unregulated supply (the 9 V battery in your case).  A switching power supply doesn't make more power, only coverts it to a different voltage x current tradeoff.  Let's say the switcher is 90% efficient.  That means the 9 V source needs to be able to supply 15 W / 90% = 16.7 W.  At 9 V that is 1.9 A.  A ordinary "9 V" battery can't do that.

A better alternative for this kind of power might be a 12 V lead-acid battery.  There are many switcher chips that can handle that voltage and make 5 V out.  The 3 A output requirement will eliminate a bunch of them, but these things are still available, especially if you are willing to use a external switching element.  Look around at offerings from ST, TI, and even Microchip at that voltage.  Linear probably has some offerings too, but they tend to be more pricy.