I like the way you asked to avoid. Forgive me if I describe it in polytope rather than fan language.

Step 1: ${\mathbb P}^1 \times {\mathbb P}^1$'s polytope is a square (or any rectangle). The four edges, taken clockwise, correspond to the ${\mathbb P}^1$s giving the classes $h_1,h_2,-h_1,-h_2$ Michael mentions.

I can only guess that by "tautological line bundle on ${\mathbb P}^1 \times {\mathbb P}^1$ you mean ${\mathcal O}(-1) \boxtimes {\mathcal O}(-1)$.

If we blow down that ${\mathbb P}^1 \times {\mathbb P}^1$, we get the affine cone over the Segre embedding of ${\mathbb P}^1 \times {\mathbb P}^1$. The polyhedron of that is also a cone, on a square.

Step 2: Blow the singular point back up, which corresponds to cutting the corner off that cone, leaving a square. So far we have an unbounded polytope that retracts to the square, just as the line bundle retracts to ${\mathbb P}^1 \times {\mathbb P}^1$.

Step 3: Projectively complete. This corresponds to bounding the cone. Combinatorially, we now have a square-based pyramid with the top corner cut off, so there's a big square on the bottom (whose class is Michael's $h$) and a little square on the top.

Step 4: Take the anticanonical class. On any toric variety, the boundary of the polytope defines an anticanonical divisor.

So far our anticanonical class is the bottom square $h$ plus the top square plus the other four faces. To calculate the linear relations between them, one needs to be precise about the locations of the vertices. I have the bottom square at $(0,0), (2,0), (0,2), (2,2)$ with $z=0$ and the top one at $(0,0), (1,0), (0,1), (1,1)$ with $z=1$. The Danilov relations from the $z$-axis vector says $$ (-1) \text{bottom} + (+1) \text{top} + 0 \text{west} + 0 \text{south} + (+1) \text{north} + (+1)\text{east} = 0 $$ so the total of the faces is $2\text{bottom} + \text{south} + \text{west}$, matching Michael's $2h+h_1+h_2$.

(As it ought, since I learned at least some of this from him.)

I like the way you asked to avoid. Forgive me if I describe it in polytope rather than fan language.

Step 1: ${\mathbb P}^1 \times {\mathbb P}^1$'s polytope is a square (or any rectangle). The four edges, taken clockwise, correspond to the ${\mathbb P}^1$s giving the classes $h_1,h_2,h_1,h_2$ Michael mentions. (EDIT: I had signs there before, by overthinking the Danilov relations.)

I can only guess that by "tautological line bundle on ${\mathbb P}^1 \times {\mathbb P}^1$ you mean ${\mathcal O}(-1) \boxtimes {\mathcal O}(-1)$.

If we blow down that ${\mathbb P}^1 \times {\mathbb P}^1$, we get the affine cone over the Segre embedding of ${\mathbb P}^1 \times {\mathbb P}^1$. The polyhedron of that is also a cone, on a square.

Step 2: Blow the singular point back up, which corresponds to cutting the corner off that cone, leaving a square. So far we have an unbounded polytope that retracts to the square, just as the line bundle retracts to ${\mathbb P}^1 \times {\mathbb P}^1$.

Step 3: Projectively complete. This corresponds to bounding the cone. Combinatorially, we now have a square-based pyramid with the top corner cut off, so there's a big square on the bottom (whose class is Michael's $h$) and a little square on the top.

Step 4: Take the anticanonical class. On any toric variety, the boundary of the polytope defines an anticanonical divisor.

So far our anticanonical class is the bottom square $h$ plus the top square plus the other four faces. To calculate the linear relations between them, one needs to be precise about the locations of the vertices. I have the bottom square at $(0,0), (2,0), (0,2), (2,2)$ with $z=0$ and the top one at $(0,0), (1,0), (0,1), (1,1)$ with $z=1$. The Danilov relations from the $z$-axis vector says $$ (-1) \text{bottom} + (+1) \text{top} + 0 \text{west} + 0 \text{south} + (+1) \text{north} + (+1)\text{east} = 0 $$ so the total of the faces is $2\text{bottom} + \text{south} + \text{west}$, matching Michael's $2h+h_1+h_2$.

(As it ought, since I learned at least some of this from him.)