Will I make it out in time?

Snails, 15 bytes

\Oo!{.,fee7.,\F

Try it online!

1 means survival while 0 means death.

Since it is impossible to outrun the fire, it is never useful to try to go around it. The best route is always a straight line. So there are only four possible choices of escape route. To determine if a direction is safe, we check for any F in the "fire cone" pointing in that direction.

Python 2, 283 218 209 208 bytes

lambda F:f(F)&f(F[::-1])
def f(F):l=F.split();w=len(l[0])+1;i=F.index('O');x,y=i/w,i%w;r=range(len(l));return all('F'in''.join(n)for n in[[l[i][x+abs(i-y):]for i in r],[l[i][max(0,y+x-i):i+x-y+1]for i in r]])

Try it online!

Takes input as a newlines separated string, and returns True/False for Dead/Alive

Works by checking each direction(udlr) for Fire in by looking outward:

Example:

Input:

FFFFF
.....
..O..
.....

Fire checks:

Up:       Down:     Left:     Right:

FFFFF               F             F
 ...                ..           ..
  O         O       ..O         O..
           ...      ..           ..

If all directions contain fire you die, otherwise there is an escape.

Edit: Back to taking a string as input, and now only checks for up/right, but also checks the input backwards (giving down/left)

Saved a lot of bytes thanks to Mr. Xcoder and Felipe Nardi Batista

JavaScript, 174 bytes

a=>+(t=>g=a=>t--?g(a.map((l,y)=>l.map((c,x)=>(h=v=>[(a[y-1]||[])[x],(a[y+1]||[])[x],a[y][x+1],a[y][x-1],c].includes(v),!c&&h()?p=1:[2,0,1].find(h))))):p)((p=a+'!').length)(a)

Input format:

• Array of Array of Integers
• 2 for F, 1 for ., 0 for O

Output:

• Truthy value (1) for survive
• Falsy value (NaN) for die

Try It:

f=a=>+(t=>g=a=>t--?g(a.map((l,y)=>l.map((c,x)=>(h=v=>[(a[y-1]||[])[x],(a[y+1]||[])[x],a[y][x+1],a[y][x-1],c].includes(v),!c&&h()?p=1:[2,0,1].find(h))))):p)((p=a+'!').length)(a)
t=s=>f(s.trim().split('\n').map(x=>x.split('').map(n=>({F:2,'.':1,O:0}[n]))))

console.log(t(`
FFFFF
.....
..O..
.....
`))

console.log(t(`
FFFF
FFFO
FFFF
`))

console.log(t(`
.F....
......
......
.F....
..O...
.FF...
.F....
..FF..
`))

console.log(t(`
...F...F
F.......
........
.F......
....O...
...F....
........
.F....F.
`))

console.log(t(`
FFF
FOF
FFF
`))

console.log(t(`
F.F
.O.
F.F
`))

console.log(t(`
....F
.....
..O..
.....
F....
`))

console.log(t(`
.F....F.
........
........
F..O....
........
.....F..
`))

console.log(t(`
...F...F
F......F
........
.F......
....O...
...F....
........
.F....F.
`))

Consider a cellular automaton. There are 3 states for a cell O (reachable by people), F (catch fired), . (nothing just happened). The rule for create next generation is:

for each cell:
  me and my 4 neighborhoods,
    if anyone is `F` then result is `F`,
    otherwise, if anyone is `O` then result is `O`
    otherwise, keep state `.`

Once there is an cell on edge has O state, the people survive. If this not happened in enough amount generation, then the people died.

// check for all neighbors:
h=v=>[(a[y-1]||[])[x],(a[y+1]||[])[x],a[y][x+1],a[y][x-1],c].includes(v)
// if me == 'O' and i'm edge (neighbors contain _undefined_), then survive
!c&&h()?p=1
// Otherwise apply the given rule
:[2,0,1].find(h)