Let us think $S^1=\{z\in\mathbb{C}:\ |z|=1\}$ and, more in general, $S^{2k+1}=\{z\in\mathbb{C}^{k+1}:\ |z|=1\}$. We consider two actions of $S^1\times S^1$:
(1) On $\mathbb{C} P^1$ via $(\theta,\zeta)[a:b]=[\theta a:\zeta b]$
(2) On $S^{2k+1}\times S^{2k+1}$ via $(\theta,\zeta)(s_1,s_2)=(\bar{\theta}s_1,\bar{\zeta}s_2)$
Then we can consider the fibred product $\mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})$, which is a fibre bundle over $\mathbb{C}P^k\times\mathbb{C}P^k$ with fibres diffeomorphic to $\mathbb{C}P^1$.
Let $\pi_1,\pi_2:\mathbb{C}P^k\times\mathbb{C}P^k\rightarrow\mathbb{C}P^k$ be the projections and let $\mathcal{O}_1(-1)=\pi_1^*\mathcal{O}_{\mathbb{C}P^k}(-1)$, $\mathcal{O}_2(-1)=\pi_2^*\mathcal{O}_{\mathbb{C}P^k}(-1)$ be the pull-backs of the tautological bundle over $\mathbb{C}P^k$. Then there is an identification $$ \begin{array}{ccc} \mathbb{P}(\mathcal{O}_1(-1)\oplus\mathcal{O}_2(-1)) &\longrightarrow & \mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})\\ [as_1,bs_2]&\longmapsto&[[a:b],(s_1,s_2)] \end{array} $$ where $[as_1,bs_2]$ is a generic element of the fibre over $([s_1],[s_2])\in\mathbb{C}P^k\times\mathbb{C}P^k$, which is $\mathbb{P}(\langle s_1\rangle\oplus\langle s_2\rangle)$.
Finally consider the tautological bundle $$ \rho:\mathcal{O}(-1)\longrightarrow\mathbb{P}(\mathcal{O}_1(-1)\oplus\mathcal{O}_2(-1)) $$ and let $u\in H^2(\mathbb{P}(\mathcal{O}_1(-1)\oplus\mathcal{O}_2(-1)))\simeq H^2(\mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1}))$ be its Euler class. If $\sigma$ is a smooth section of $\rho$ transverse to the zero-section, then $PD(u)$ is represented by the submanifold $\sigma^{-1}(0)$.
Question: what is an example of such $\sigma$?
I am aware that finding $\sigma$ tantamounts to finding a complex number $c(a,b,s_1,s_2)$ depending on $a,b,s_1,s_2$ such that $$ c(\lambda\theta a,\lambda\zeta b,\bar{\theta}s_1,\bar{\zeta}s_2)=\frac{1}{\lambda}c(a,b,s_1,s_2) $$ for any $\lambda\in\mathbb{C}\setminus\{0\}$ and $(\theta,\zeta)\in S^1\times S^1$, but I am not able to find it. Any hints?
N.B.: I am reasonably happy if $\sigma$ is not smooth but just meromorphic since then I can find $PD(u)$ in terms of the zeroes and poles of $\sigma$ anyway.
Addendum:
So it seems that a possible (meromorphic) choice is $$ \sigma([as_1,bs_2])=\left(\frac{as_1}{bs_2^0},\frac{s_2}{s_2^0}\right), $$ where $s_2=(s_2^0,\ldots,s_2^k)$ as an element of $\mathbb{C}^{k+1}$.
If I am not wrong (am I?) it looks like $\sigma$ has no zeroes and that it has poles when either $b=0$ or when $a\neq0,s_2^0=0$. What about the case $a=0,s_2^0=0$?
Question: Can any one confirm I am describing the zeroes and the poles of $\sigma$ correctly?
Question: Is there the possibility of finding a smooth section rather than meromorphic?
Follow-up:
Thanks to Jez answers now we know that $\sigma$ is a right choice. In case anyone is interested I will keep explaining my problem and tell the next point where I am stuck:
Using the identification $$ \mathbb{P}(\mathcal{O}_1(-1)\oplus\mathcal{O}_2(-1))\simeq \mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1}) $$ above it turns out that the pole locus of $\sigma$ is the union of $$ P:=\{[1:0]\}\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1}) $$ and $$ Q:=\mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times (S^{2k+1}\cap H^0)) $$ where $H^0=\{(z^0,\ldots,z^k)\in\mathbb{C}^{k+1}:\ z^0=0\}$. Then $$ PD(u)=-[P]-[Q]. $$ For brevity let us define $\alpha:=PD^{-1}[P]$ and $\beta:=PD^{-1}[Q]$. Hence $u=-\alpha-\beta$.
Given the map $$ \begin{array}{rccc} \iota: & \mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})& \longrightarrow & (\mathbb{C}P^1\times\mathbb{C}P^1)\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})\\ & \left[x,(s_1,s_2)\right] & \longmapsto & [(x,x),(s_1,s_2)] \end{array} $$ I would like to compute $\iota^!(u)=-\iota^!(\alpha)-\iota^!(\beta)$. Note that $$ \iota^!(\alpha)=(PD^{-1}\circ\iota_\ast\circ PD)(\alpha)=(PD^{-1}\circ\iota_\ast)[P]=PD^{-1}[\iota(P)]. $$ Similarly $\iota^!(\beta)=PD^{-1}[\iota(Q)]$.
I know how to prove (I can provide the details if needed) that $\iota^!(\alpha)=\rho_1^*\alpha\smile\rho_2^*\alpha$ where $$ \begin{array}{rccc} \rho_j: & (\mathbb{C}P^1\times\mathbb{C}P^1)\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})& \longrightarrow & \mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})\\ & \left[(x_1,x_2),(s_1,s_2)\right] & \longmapsto & [x_j,(s_1,s_2)] \end{array} $$
Question: What is $\iota^!(\beta)$?
It is quite straightforward to prove that $\beta=-\pi^*c_1(\mathcal{O}_2(-1))$, where $$ \pi:\mathbb{C}P^1\times_{S^1\times S^1}(S^{2k+1}\times S^{2k+1})\longrightarrow\mathbb{C}P^k\times\mathbb{C}P^k $$ is the projection and also that $$ \iota(Q)=\Delta_{\mathbb{C}P^1\times\mathbb{C}P^1}\times_{S^1\times S^1}(S^{2k+1}\times (S^{2k+1}\cap H^0)) $$ but I do not know more than that yet.
I believe your section $\sigma$ is correct, except that the denominator in the first term should be $b s^0_1$ instead of $b s^0_2$. This has poles on $\{b=0\} \cup \{s^0_1=0\} \cup \{s^0_2=0\}$. If it were finite at $a=s^0_1=0$ then by continuity it would have to be finite at $a=\varepsilon, s^0_1=0$ for small $\varepsilon$, which it isn't, so $a=s^0_1=0$ is a pole.
To answer your other questions it really comes down to the distinction between smooth and holomorphic. There is a smooth section transverse to the zero section: simply use a partition of unity to perturb the zero section.
There is, however, no holomorphic global section except $0$. To see this, note that we have a tower of bundles: $E \rightarrow P \rightarrow \mathbb{CP}^k\times \mathbb{CP}^k$, where $P$ is the projective bundle and $E$ is the tatuological bundle over $P$, and suppose $\sigma$ is a holomorphic global section of $E$ over $P$. For each $(s_1, s_2) \in \mathbb{CP}^k \times \mathbb{CP}^k$ consider restricting $\sigma$ to the fibre of $P$ over $(s_1, s_2)$. This gives a holomorphic section of $\mathcal{O}_{\mathbb{CP}^1}(-1)$, which must therefore be $0$. Hence $\sigma$ vanishes on all of $P$.
Added in response to comment:
The statement above about the denominator is incorrect; the version given in the question is right (and a corresponding comment applies to the pole locus). Explicitly, on $\{b, s^i_2 \neq 0\}$ we can choose our trivialisation $\phi_i$ of $E$ so that that the $i$-component in the second factor is $1$. Then the transition map $\psi_{ij}$ is given by $s^j_2/s^i_2$.
I had been uneasy about the apparent asymmetry in the pole locus $\{b=0\} \cup \{s^0_2=0\}$ suggested in the question, given that the whole setup is invariant under transposing the factors, but of course if you make this transposition then not only do $s^0_1$ and $s^0_2$ get swapped, but $a$ and $b$ too.