What you say is correct and here is a proof.
First remark is that if there were such an action of $T^2$ holomorphic on the fibers and smooth on $M$, it would be just simply holomorphic on $M$. Indeed, for a fixed toric surface, $T^2$ can act on it holomorphically only in a discreet number of ways, that (provided we fix a base in $H_1(T^2,\mathbb Z))$ are parametrized by $SL(2,\mathbb Z)$. The space $M\setminus M_0$ is a locally holomorphically trivial fibration, so $T^2$ would act on it holomorphically. Now, since the action of $T^2$ on $M$ is continuous it must be holomorphic, because any continuous vector field on a complex manifold, holomorphic outside a set of co-dimension $1$ is holomorphic.
So, to finish we need to show that there is no holomorphic action of $T^2$ on $M$. To do this take first any Kahler metric on $M$ and homogenise it under the action of $T^2$. Now, the action of $T^2$ becomes Hamiltonian, so we have a fiberwise moment map. For each fiber the image of the moment map is a $4$-gon on $\mathbb R^2$ and this $4$-gon varies continuously with the fiber. Here we get a contradiction. Indeed, the polygons corresponding to $F_n$ can not be continuously deformed to those of $F_{n-2m}$ without degenerating them at some point (I assume $n\ne m$ but I guess one can exclude this case as well).
Note that in the last reasoning we could have just taken a product Kahler form on $\mathbb P^1\times \mathbb P^2\times \mathbb C$ and restrict it to $M$. In this case the image of the moment map would not have varied at all (but the above reasoning is more universal since it does not use any information about $M$).
The first thing I would like to point out is that the variety we care about is actually
{$([x_0 , x_1 ],[y_0,y_1,y_2]):{x_0}^n y_1 - {x_1}^n y_2 =0$}$ =V \subset \mathbb{P}^1 \times \mathbb{P}^2$
(the defining equations you gave are not usually homogenous, and this is what Huybrechts gives in his book: http://books.google.com/books?id=eZPCfJlHkXMC&q=hirzebruch).
We want to start out by thinking about $\mathbb{P}^1 \times \mathbb{P}^2$ as a $\mathbb{P}^2$ bundle over $\mathbb{P}^1$. In particular, it comes with the distinguished line bundle $\pi_2^*\mathscr{O}_{\mathbb{P}^2}(1)$ ($\pi_2$ is the projection onto $\mathbb{P}^2$). Now we examine the short exact sequence of sheaves of modules which defines V:
$0\rightarrow \pi_1^*\mathscr{O}_{\mathbb{P}^1}(-n) \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(-1)\rightarrow \mathscr{O} _{\mathbb{P}^2\times\mathbb{P}^1} \rightarrow \mathscr{O}_V\rightarrow0$.
We would like to apply ${\pi_1}_*$ on this sequence after twisting by $\pi_2^*\mathscr{O}_{\mathbb{P}^2}(1)$. The point of this is that if you believe that $V$ is a $\mathbb{P}^1$ bundle over $\mathbb{P}^1$ (which you should because you can check it on charts) then ${\pi_1}_*(\mathscr{O}_V \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(1))$ should be the corresponding locally free sheaf of rank 2. So we can then check to see if it is isomorphic to $\mathscr{O} _{\mathbb{P}^1}\oplus\mathscr{O} _{\mathbb{P}^1}(n)$.
After twisting and applying ${\pi_1}_*$, we get the sequence:
$0\rightarrow \mathscr{O}_{\mathbb{P}^1}(-n)\rightarrow \mathscr{O} _{\mathbb{P}^1}^{\oplus 3} \rightarrow {\pi_1}_*(\mathscr{O}_V \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(1))\rightarrow0$.
(In general one does not get exactness on the right, but in a situation like this we do)
And by figuring out what our maps are doing we should be able to argue that the sheaf on the right is some line bundle twist of $\mathscr{O}_ {\mathbb{P}^1}\oplus \mathscr{O}_{\mathbb{P}^1} (n)$, which implies $V \cong \mathbb{P}(\mathscr{O} _{\mathbb{P}^1}\oplus\mathscr{O} _{\mathbb{P}^1}(n))$.
Best Answer
You can take $y_0 = 0$, $y_1 = -tx_1^m$, $y_2 = x_0^m$.
EDIT. Here is a simple computation of the normal bundle of the curve $$ C_t = \{([x_0:x_1],[0:-tx_1^m:x_0^m])\} \subset \mathbb{P}^1 \times \mathbb{P}^2 $$ in the surface $$ S_t = \{x_0^ny_1−x_1^ny_0+tx_0^{n−m}x_1^my_2=0\} $$ for $t \ne 0$. First, the normal bundle fits into the exact sequence $$ 0 \to N_{C_t/S_t} \to N_{C_t/\mathbb{P}^1 \times \mathbb{P}^2} \to N_{S_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert_{C_t} \to 0. $$ Since $C_t$ is a graph of a map $\mathbb{P}^1 \to \mathbb{P}^2$ (of degree $m$), its normal bundle is the pullback of the tangent bundle of $\mathbb{P}^2$, hence its degree is $$ \deg(N_{C_t/\mathbb{P}^1 \times \mathbb{P}^2}) = 3m. $$ On the other hand, $S_t$ is a divisor of type $(n,1)$ on $\mathbb{P}^1 \times \mathbb{P}^2$, hence its normal bundle restricted to $C_t$ has the degree equation which has degree $$ \deg(N_{S_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert_{C_t}) = n + m. $$ Therefore, $$ \deg(N_{C_t/S_t}) = 3m - (n + m) = 2m - n. $$