Following Mike Roth's correct comment, I would just like to have a correct answer on record. This is one of a series of examples that I learned of from Tom Graber, but which I guess goes back to the work on "normic forms". Assume that the characteristic is not $3$ (there are similar examples in every characteristic). Let $\mathbb{P}^1$ have homogeneous coordinates $[U,V]$. Let $\mathbb{P}^3$ have homogeneous coordinates $[T_0,T_1,T_2]$. Consider the closed subscheme $Y$ of $\mathbb{P}^1\times \mathbb{P}^2$ with bihomogeneous defining equation $$ F(U,V;T_0,T_1,T_2) = U^2T_0^3 + UVT_1^3 + V^2T_2^3.$$ There is an action of $\mathbb{G}_m$ on $\mathbb{P}^1\times \mathbb{P}^2$ where $\lambda\in \mathbb{G}_m$ acts by $$ \lambda \bullet ([U,V],[T_0,T_1,T_2]) = ([U,\lambda^{-3}V],[T_0,\lambda T_1,\lambda^2 T_2]).$$ The homogeneous polynomial $F$ is invariant for this (bilinearized) action. Thus $Y$ is invariant.
Of course $Y$ is singular at the points $([1,0],[0,0,1])$ and $([0,1],[1,0,0])$, but that changes nothing. There exists a resolution $\nu:X\to Y$ that is projective (just a sequence of blowings up). Since $X$ is a smooth, projective variety of dimension $2$, since $\mathbb{P}^1$ is a smooth, projective variety of dimension $1$, and since the morphism $f := \text{pr}_{\mathbb{P}^1}\circ \nu$ is surjective, automatically $f:X\to \mathbb{P}^1$ is flat and projective.
There is certainly a $\mathbb{P}^1$-flat divisor in $X$: just take the closure in $X$ of any positive degree, effective zero cycle on the generic fiber, e.g., the common zero locus of $F$ and $G=T_0+T_1+T_2$. On the other hand, I claim that there is no rational section of $f$.
If there were a rational section of $f$, then its image in $Y$ would be a rational section of $\text{pr}_{\mathbb{P}^1}:Y\to \mathbb{P}^1$. The Zariski closure of this rational section would be a curve, which then gives a point of the Hilbert scheme parameterizing curves on $Y$. The action of $\mathbb{G}_m$ on $Y$ induces an action of $\mathbb{G}_m$ on the Hilbert scheme. Consider the orbit under $\mathbb{G}_m$ of the specified point of the Hilbert scheme. Since the (connected components of the) Hilbert scheme are projective, the closure of this orbit is proper. In particular, there exists a "limit at infinity" of the original Hilbert point.
This "limit point" parameterizes a curve $C$ in $Y$ that is a limit of rational sections, and that is $\mathbb{G}_m$-invariant. By the valuative criterion of properness applied to $f$, every limit of a one-parameter family of rational sections of $f$ is, again, a rational section of $f$, i.e., there is a unique component of $C$ that dominates $\mathbb{P}^1$, and this component is the image of a rational section of $f$. Since $C$ is $\mathbb{G}_m$-invariant, also the rational section is $\mathbb{G}_m$-equivariant.
However, the only $\mathbb{G}_m$-equivariant rational sections of the projection, $$\text{pr}_{\mathbb{P}^1}:\mathbb{P}^1\times \mathbb{P}^2\to \mathbb{P}^1,$$ are "monomial" sections, i.e., $$ h([U,V]) = ([U,V],[c_0U^{a_0}V^{b_0},c_1U^{a_1}V^{b_1}, c_2U^{a_2}V^{b_2}]), $$ where $c_0$, $c_1$, $c_2$ are elements in the ground field such that $(c_0,c_1,c_2)\neq (0,0,0)$, and where each $a_i$ and $b_i$ is a positive integer such that $a_0+b_0=a_1+b_1=a_2+b_2 = e$ for some positive integer $e$. But then the restriction of the equation $F$ on this section is, $$ F\circ h([U,V]) = c_0^3U^{3a_0+2}V^{3b_0} + c_1^3U^{3a_1+1}V^{3b_1+1} + c_2^3U^{3a_2}V^{3b_2+2}. $$ In particular, the congruence classes modulo $3$ of the exponent vectors of the $3$ terms are $(\overline{2},\overline{0})$, $(\overline{1},\overline{1})$, and $(\overline{0},\overline{2})$. So there can be no cancellation, i.e., the monomials are linearly independent in $k[U,V]$. So the only way that this linear combination of monomials may be zero is if $(c_0^3,c_1^3,c_2^3)$ equals $(0,0,0)$, contradicting that $(c_0,c_1,c_2)\neq (0,0,0)$. This contradiction proves that there is no rational section of $f$.
By the way, this works more generally to show that for every triple of positive integers $(r,d,n)$ with $n+1=d^r$, there exists a degree $d$ hypersurface $Y$ in $\mathbb{P}^r\times \mathbb{P}^n$ such that the projection $Y\to \mathbb{P}^r$ admits no rational section. These are "normic forms" showing that the Tsen-Lang theorem is sharp.