Let $X$ be a smooth projective variety. A Higgs bundle is a pair $(E, \phi)$, where $E$ is a vector bundle, $\phi: E \to \Omega_X \otimes E$ is a $\mathcal{O}_X$-linear map s.t. $\phi \wedge \phi = 0$. Map $\phi$ is called the Higgs field.
Let $\pi:|\Omega_X|\to X$ be the total of the vector bundle $\Omega_X$, vector bundles over $|\Omega_X|$ are the same as vector bundles over $X$ with a map $\pi_*(\mathcal{O}_{|\Omega_X|}) \otimes E \to E$. Using $\pi_*(\mathcal{O}_{|\Omega_X|}) \cong \operatorname{Sym} T_X$ we can say that any map $\pi_*(\mathcal{O}_{|\Omega_X|}) \otimes E \to E$ is determined by its first degree part: $T_X \otimes E \to E$.
Higgs field $\phi: E \to \Omega_X \otimes E$ is equivalent to the map $T_X \otimes E \to E$, so by Higgs field any Higgs bundle is a bundle on $|\Omega_X|$. What is the meaning of condition $\phi \wedge \phi=0$ for this vector bundle on $|\Omega_X|$?