First note that $M = \overline{\mathbb F_p} (x)$ because that field contains $t$ (it's $f/x^i$), so contains $\overline{\mathbb F_p}(t)$, and is generated over it by $x$, which is a root of $f -t x^i=0$.

1) This is correct:

You can observe that all the places lying over $t=0$ had better be places of $M$, which correspond to points of the projective line. If they are not the point at $\infty$, they lie over $t=0$ if and only if they are roots of $f$ (zero would be special here if we didn't get rid of it), and their ramification index is their multiplicity in $f$, as one can see by examining $f/x^i$, which is just $f$ in this case, in local coordinates.

The point $x=\infty$ is mapped to $t=\infty$ if $\deg f>i$, $t=0$ if $\deg f<i$, and something else if they are equal. In particular if $i=0$ it's mapped to $\infty$, so does not contribute to the ramification over $t=0$.

2) The only places that can possibly be sent to the infinite place are the pole $0$ of $f/t^i$, and the point $\infty$. As mentioned, $\infty$ is mapped to the infinite place if $i < \deg f$. The ramification index is again the order of vanishing of a local coordinate at $t =\infty$. We can take the local coodrinate to be $t^{-1}$, so we are interested in the order of vanishing of $x^i/f(x)$. The order of vanishing at $x=0$ is clearly $i$, unless $x$ happens to divide $f$, in which case it is lesser, so that is the ramification index at $0$. To get the order of vanishing at $\infty$, we need to choose a local coordinate at $\infty$, say $y= x^{-1}$, obtaining $ y^{-i} / f(y^{-1})$. We can factor $f(y^{-1})$ as $y^{-\deg f}$ times a polynomial in $y$ nonvanishing at $y=0$, so this is $y^{\deg f-i}$ divided by a polynomial in $y$ nonvanishing at $y=0$, giving a ramification index of $\deg f-i$.

So there are two ramification points, one of index $i$ and one of index $\deg f-i$, with the latter point being removed if the formula for the index is not positive, and the former point having its index reduced if $x$ divides $f$.

3) The assumption can't change anything because it is a special case, but it does put very strong assumptions on $f$. In fact it should not be too hard to classify all $f$ such that this map is Galois.