Let $X$ be a smooth projective variety of dimension $n$. Take the bundle $TX \oplus Sym^2(TX)$ over $X$ where $Sym^2(TX)$ is the second symmetric product of the tangent space. 
The Grassmannian bundle $Gr(n,TX \oplus Sym^2(TX))$ has a canonical section, namely $TX$. 


My question is: what is the Poincare dual of this section in the cohomology ring of the Grassmannian bundle? 


The cohomology ring is


$H^\*(Gr(n,TX \oplus Sym^2(TX)))=H(X)[c\_1,\ldots, c\_n,d\_1,\ldots, d\_{{n+1 \choose 2}}]/$


$(1+c\_1+\ldots +c\_n)(1+d\_1+\ldots +d\_{{n+1 \choose 2}})=c(TX \oplus Sym^2(TX))$


This is probably a trivial question I am a bit confused about it now. 



Sections of Grassmannian bundles
Let $U$ denote the tautological subbundle. I guess that $c\_i = c\_i(U^\*)$ in your notation. Consider the composition map
$$
U \to p^\*(TX + S^2TX) \to p^\*S^2TX,
$$
where $p:Gr \to X$ is the projection. Then your section is the zero locus of this map. In other words, it is the zero locus of a global section of the vector bundle $U^\*\otimes p^\* S^2TX$. The section is regular, so the class of the zero locus equals the top Chern class of the bundle. Thus the answer is
$$
c\_{n^2(n+1)/2}(U^\*\otimes p^\*S^2TX).
$$
The rest is a straightforward computation.


