You have already done the case, when $i = 1$.

Now suppose it i true for $i$. Let's prove it for $i+1$. Suppose $H \triangleleft G$ has Fitting length $i+1$. That means $\frac{H}{U_iH}$ is nilpotent. Now let's define $\phi_i$ as a natural homomorphism between $G$ and $\frac{G}{U_iG}$. Then $\phi_i(H)$ is a nilpotent normal subgroup of $\frac{G}{U_iG}$ because $U_iH$ lies in $U_iG$ as a normal subgroup (characteristic subgroup of a normal subgroup is always normal) of Fitting length $i$ by induction step. That means, that $\phi_i(H) \triangleleft U_1\frac{G}{U_iG}$, which results in $H \triangleleft \phi_i^{-1}(U_1\frac{G}{U_iG})=U_{i+1}G$.

Thus, $\forall i \in \mathbb{N}$ every normal subgroup of $G$ of fitting length $i$ is contained in $U_iG$.