Combined signal folding and shifting
The statement $g[n]=f[-n]$ is meaningless unless you assign some numerical value to $n$, e.g. $g[3]=f[-3]$, and if you do say that $n=3$, then the equality
$g[3] = f[-3]$ tells you nothing about the relationship between $g[n+k]=g[3+k]$
to the value taken on $f[\cdot]$ for any choice of argument.

What you really meant to say is that


  $$\text{For all integers } n, g[n]=f[-n]$$


which asserts that the equality holds for all choices of $n$, and in
this case, it follows by definition that $g[n+k] = f[-n-k]$;
you just need to negate the argument $n+k$ of $g$ (to get $-n-k$)
and use it (it meaning $-n-k$) as the argument of $f$ exactly as 
the definition says you should.  The 
complicated arguments in another answer that claim that $g[n+k]=f[-n+k]$
are wrong, as discussed in the comments following that answer.