Is there a way to take a battery power source at 7.1volts 15Amps and convert it to 5volts 4Amps?
First, when you see a power supply rated at \$V\$ volts and \$A\$ amps, it usually means that it'll produce \$V\$ volts at up to \$A\$ amps.  If you present that supply with a device that'll take \$V\$ volts and consume \$A / 3\$ amps -- that's what'll happen.  For any given load presented with any given voltage, it'll take what current it takes.  A source cannot force a voltage and current on a load -- only one or the other.
Moreover, when a battery pack is marked \$V\$ volts, it's just understood that the rating is nominal.  For batteries, that's usually the voltage at the middle of their useful discharge curve.  So your &quot;7.1V&quot; battery pack will put out 8V or more when it's freshly charged, and as low as 6.4&quot; when it's discharged.  Moreover, Li-anything batteries (Li-Ion, Li-polymer, Li-Fe, etc.) pretty much kill themselves when you discharge them too far, so you need to use a circuit that has a low-voltage cutoff, or you need to lay in a supply of extra cells.  You also need to be careful if through accident you do over-discharge them, because unprotected cells that have been discharged too far can catch fire when you recharge them.
Second, there's an infinite number of ways you can skin this cat, but there's three that I'm going to mention:

Obvious, but don't do this: use a resistor that drops 2.1V at 4A.  This seems like a good idea, but both the battery voltage and the load current will vary, so you'll get just about anything but 5V at the output.
Electrically easy, but thermally hard: use a 5V linear regulator rated for 5A, like a 7805, and a big heat sink.  Linear regulators work by passing the load current and dropping voltage through the device, which basically means they burn up a bunch of power as a price of electrical simplicity.  So with an 8V battery your regulator will need to dissipate 12W -- this will require a big heat sink.
A bit harder, with pitfalls, but efficient and cool (in the sense that you can touch bits of the circuit and not get burnt): Use a buck converter as @Hearth mentions.  You can buy cheap no-name modules for this off of eBay, you can buy more trusty modules for this from distributors like DigiKey, Mouser, etc., in the US or Farnell in Europe.  The cheap modules tend to just copy the datasheet circuit, and don't always do the job.

You mention a fan -- this can be problematical, because typically a circuit with a motor in it that's rated for a current and voltage will give you the current that the motor will need once it is spinning.  There may be some &quot;inrush&quot; current, possibly much higher, needed to actually start the motor.  I say &quot;may&quot; because one could design a fan circuit to limit the inrush current -- I just have little faith that it's actually done.
So I recommend that, first, you have a separate buck converter for your &quot;clean&quot; 5V supplies (your Arduino and any other logic) and your &quot;dirty&quot; 5V supplies (your fan module).  Then I recommend that you read the fan module's datasheet carefully for inrush current.  Batteries can easily handle brief overloads (i.e., a battery rated for 4A could do 12A for 1/10th of a second, no problem).  Buck converters can't (unless &quot;brief&quot; means 10 milliseconds, and maybe not then).  So your &quot;dirty&quot; 5V supply needs to be sized for the inrush current of any motors &amp; whatnot.
And shop carefully for your module -- I can't tell you who's reliable beyond CUI and TI, and I don't even know if CUI makes modules.  But if you get something cheap from a surplus place, an auction house, or someone that also sells dishware, you'll be disappointed.