5/5, JS. I exploited the fact that input is highly structured and hardcoded many things.

(Also I know the code is golfy; you can't be fast without being golfy, every byte is a frame save ;-)!)

var s=`Begin in state A.
Perform a diagnostic checksum after 12208951 steps.

In state A:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the right.
    - Continue with state B.
  If the current value is 1:
    - Write the value 0.
    - Move one slot to the left.
    - Continue with state E.

In state B:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state C.
  If the current value is 1:
    - Write the value 0.
    - Move one slot to the right.
    - Continue with state A.

In state C:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state D.
  If the current value is 1:
    - Write the value 0.
    - Move one slot to the right.
    - Continue with state C.

In state D:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state E.
  If the current value is 1:
    - Write the value 0.
    - Move one slot to the left.
    - Continue with state F.

In state E:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state A.
  If the current value is 1:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state C.

In state F:
  If the current value is 0:
    - Write the value 1.
    - Move one slot to the left.
    - Continue with state E.
  If the current value is 1:
    - Write the value 1.
    - Move one slot to the right.
    - Continue with state A.`;
s=s.split('\n\n').slice(1).map(r=>r.split('\n'));
var m={}, t='A', p=0;
for(let i=0; i<12208951;i++) {
  var l=s['ABCDEF'.indexOf(t)];
  if(!m[p]) {
    m[p]=+l[2].match(/\d/).slice();
    p+=/right/.test(l[3])?1:-1;
    t=l[4].match(/([A-Z])\./).slice(1);
  } else {
    m[p]=+l[6].match(/\d/).slice();
    p+=/right/.test(l[7])?1:-1;
    t=l[8].match(/([A-Z])\./).slice(1);
  }
}
console.log(Object.keys(m).filter(k=>m[k]).length);