Make An ASCII Poker Chip Stack Arrangement

JavaScript (ES6), 185 177 ... 171 bytes

f=(a,s='')=>(a=(a[0]?a:[...(m='25455')+m+m].map((m,i)=>(a-=(k=a/(v/=m)|0)*v,k*16+i),v=1E9).sort((a,b)=>b-a)).map(n=>n>15&&(s+='SVTILAMCOPYBGRW'[n&15],n-16)),s&&f(a)+`
`+s)

Formatted and commented

(a, s = '') => (                           // a = bankroll OR array of chip stacks
  a = (                                    // s = string of chip initials for this iteration
    a[0] ?                                 // if 'a' already is an array:
      a                                    // use it as-is
    :                                      // else, 'a' is the bankroll:
      [...(m = '25455') + m + m]           // for each 'm' in [ 2, 5, 4, 5, 5, ... ] (x3)
      .map((m, i) =>                       // v = current chip value = previous value / m
        (                                  // k = floor(a / v) = number of these chips
          a -= (k = a / (v /= m) | 0) * v, // update remaining bankroll: a = a - k * v
          k * 16 + i                       // saved in array: chip index in bits 0-3
        ),                                 // and number of chips in bits 4+
        v = 1E9                            // initial chip value = 1 billion
      )                                    // (1B / 2 -> 500M, 500M / 5 -> 100M, etc.)
      .sort((a, b) => b - a)               // sort the chip stacks
  )                                        //
  .map(n =>                                // for each chip stack:
    n > 15 &&                              // if there's at least one remaining chip of
    (                                      // this kind:
      s += 'SVTILAMCOPYBGRW'[n & 15],      // append its initial to the string
      n - 16                               // decrement the number of chips in this stack
    )                                      //
  ),                                       // process recursive call if there was at least
  s && f(a) + '\n' + s                     // one non-empty chip stack (the next lines are
)                                          // appended at the beginning of the final string)

Demo

let f=(a,s='')=>(a=(a[0]?a:[...(m='25455')+m+m].map((m,i)=>(a-=(k=a/(v/=m)|0)*v,k*16+i),v=1E9).sort((a,b)=>b-a)).map(n=>n>15&&(s+='SVTILAMCOPYBGRW'[n&15],n-16)),s&&f(a)+`
`+s)

function update() {
  document.getElementById('o').innerHTML = f(+document.getElementById('i').value);
}
update();

<input id="i" value="1342185143" oninput="update()"><pre id="o"></pre>

Ruby, 181 177 bytes

->n{*a=5*10**8
14.times{|i|a<<a[-1]/[5,4,5,5,2][i%5]}
a=a.zip("SVTILAMCOPYBGRW".chars).map{|v,c|[-(n/v),v,c,n%=v]}.sort
l=-a[0][0]
(1..l).map{|i|a.map{|d|l-i<-d[0]?d[2]:""}*""}}

Tests on Ideone.

Pyth, 56 55 52 bytes

The code contains some unprintables, so here's a reversible xxd hexdump.

00000000: 3d48 516a 5f2e 745f 2023 6c44 2a56 2e22  =HQj_.t_ #lD*V."
00000010: 4159 261c 0c24 2087 0c86 1e22 6d68 412e  AY&..$ ...."mhA.
00000020: 4448 645f 2e75 2a4e 5950 2a33 6a37 3733  DHd_.u*NYP*3j773
00000030: 3620 362f                                6 6/

Try it online. Test suite.

Pro tip: to golf 1 in the end of a program accepting a nonzero number in Q, just add /. /QQ, which in Python would be Q // Q, is 1 for nonzero Q.

Without compression and unprintables (55 bytes):

=HQj_.t_ #lD*V"SVTILAMCOPYBGRW"mhA.DHd_.u*NYP*3j7736 6/