Your question is not so clearly stated; so I take the opportunity to interpret it and write a little essay.

As far as I understand your question, you found out that the multiplicativity result by "Borel-Hirzebruch" (I think it is actually due to Hirzebruch-Chern-Serre) implies that the higher Chern character of the index bundle of the signature operator are zero and now you want to see a more conceptual explanation. You should have a look into Atiyah's "The signature of fibre bundles". The following is definitly in the spirit of that paper, though I did not find the statement there.

Theorem: "Let $Z \to X$ be an oriented smooth fibre bundle with fibre $Y$, closed, of dimension $4k$. Pick a fibrewise Riemann metric and let $D$ be the fibrewise signature operator. Assume that $\pi_1 (X)$ acts trivially on $H^{2k}(Y; \mathbb{R})$. Then the index bundle of the signature operator $ind(D)$ is trivial of virtual rank $sign(Y)$. Moreover, if the image of $\pi_1 (X)$ in $Gl(H^{2k} (Y); \mathbb{R})$ is finite, then the components of $ch(ind(D))$ in positive degree are zero."

Proof:

1. A linear algebra fact. Given a finite-dimensional real vector space $V$ with a nondegenerate symmetric bilinear form $b$; of signature $(p,q)$. Consider the space $Q(b)$ of all pairs $(V_{+},V_{-})$, such that $\pm b|_{V_{\pm}} $ is positive definite and $W_{-} \oplus W_{+}=V$. $Q$ is a subspace of the product $Gr_p (V) \times Gr_q(V)$ of two Grassmannians. $Q(b)$ is diffeomorphic to the homogeneous space $O(p,q)/(O(p)\times O(q))$, and $O(p) \times O(q) \subset O(p,q)$ is a maximal compact subgroup. Therefore $Q$ is contractible (there should be a more direct proof, though).

2. Assume that the $\pi_1$-action on $V:=H^{2k} (Z_x; \mathbb{R})$ is trivial. There is the intersection form $b$ on $V$; pick a decomposition $V=W_{+} \oplus W_{-}$ as before. Now consider the bundle $E:=H^{2k} (Z/X) \to X$; the fibre over $x$ is the cohomology $H^{2k} (Y_x; \mathbb{R})$. Because the $\pi_1$-action is trivial, the bundle is trivial and the splitting above gives a splitting $E=F_{+} \olpus F_{-}$ into two trivial subbundles.

3. Now pick a Riemann metric on the fibres. The Hodge theorem identifies $E$ with the bundle of harmonic $2k$-forms and the Hodge star $\ast$ is an involution and it splits $E= E_{+} \oplus E_{-}$ into the sum of eigenspaces. More or less by definition, $ind (D) = [E_{+}]-[E_{-}]$. Now I claim that $E_{\pm}\cong F_{\pm}$. This is because both decompositions can be viewed as sections to the bundle $X \times Q(b)$ and since $Q(b)$ is contractible, they are homotopic.

4. If the image of the monodromy is a finite group, say $\mu: \pi_1 (X) \to G$, you pass to the finite cover $\tilde{X} \to X$ corresponsing to the kernel of $\mu$. The pullback of $Z$ to $\tilde{X}$ has trivial monodromy and so the signature index bundle is trivial. But the induced homomorphism $H^{\ast} (X; \mathbb{Q}) \to H^{\ast}(\tilde{X}, \mathbb{Q})$ is injective, which allows to reduce the argument to the case of a trivial action. QED

After I have written all this, I realize that the main point is that the bundle $E$ has structural group $O(p,q)$ and it is flat as such a bundle. It allows a reduction of the structural group to $O(p) \times O(q)$ (by the contractibility of $Q(b)$), but not as a flat bundle. If you find a flat reduction, then $ch(Ind(D))=0$ (in positive degrees) by Chern-Weil theory. The index theorem is not really relevant here; it identifies $ch(ind(D))$ with the fibre integral of the $L$-class of the vertical tangent bundle.

Your question is not so clearly stated; so I take the opportunity to interpret it and write a little essay.

As far as I understand your question, you found out that the multiplicativity result by "Borel-Hirzebruch" (I think it is actually due to Hirzebruch-Chern-Serre) implies that the higher Chern character of the index bundle of the signature operator are zero and now you want to see a more conceptual explanation. You should have a look into Atiyah's "The signature of fibre bundles". The following is definitly in the spirit of that paper, though I did not find the statement there.

Theorem: "Let $Z \to X$ be an oriented smooth fibre bundle with fibre $Y$, closed, of dimension $4k$. Pick a fibrewise Riemann metric and let $D$ be the fibrewise signature operator. Assume that $\pi_1 (X)$ acts trivially on $H^{2k}(Y; \mathbb{R})$. Then the index bundle of the signature operator $ind(D)$ is trivial of virtual rank $sign(Y)$. Moreover, if the image of $\pi_1 (X)$ in $Gl(H^{2k} (Y); \mathbb{R})$ is finite, then the components of $ch(ind(D))$ in positive degree are zero."

Proof:

1. A linear algebra fact. Given a finite-dimensional real vector space $V$ with a nondegenerate symmetric bilinear form $b$; of signature $(p,q)$. Consider the space $Q(b)$ of all pairs $(V_{+},V_{-})$, such that $\pm b|_{V_{\pm}} $ is positive definite and $W_{-} \oplus W_{+}=V$. $Q$ is a subspace of the product $Gr_p (V) \times Gr_q(V)$ of two Grassmannians. $Q(b)$ is diffeomorphic to the homogeneous space $O(p,q)/(O(p)\times O(q))$, and $O(p) \times O(q) \subset O(p,q)$ is a maximal compact subgroup. Therefore $Q$ is contractible (there should be a more direct proof, though).

2. Assume that the $\pi_1$-action on $V:=H^{2k} (Z_x; \mathbb{R})$ is trivial. There is the intersection form $b$ on $V$; pick a decomposition $V=W_{+} \oplus W_{-}$ as before. Now consider the bundle $E:=H^{2k} (Z/X) \to X$; the fibre over $x$ is the cohomology $H^{2k} (Y_x; \mathbb{R})$. Because the $\pi_1$-action is trivial, the bundle is trivial and the splitting above gives a splitting $E=F_{+} \oplus F_{-}$ into two trivial subbundles.

3. Now pick a Riemann metric on the fibres. The Hodge theorem identifies $E$ with the bundle of harmonic $2k$-forms and the Hodge star $\ast$ is an involution and it splits $E= E_{+} \oplus E_{-}$ into the sum of eigenspaces. More or less by definition, $ind (D) = [E_{+}]-[E_{-}]$. Now I claim that $E_{\pm}\cong F_{\pm}$. This is because both decompositions can be viewed as sections to the bundle $X \times Q(b)$ and since $Q(b)$ is contractible, they are homotopic.

4. If the image of the monodromy is a finite group, say $\mu: \pi_1 (X) \to G$, you pass to the finite cover $\tilde{X} \to X$ corresponsing to the kernel of $\mu$. The pullback of $Z$ to $\tilde{X}$ has trivial monodromy and so the signature index bundle is trivial. But the induced homomorphism $H^{\ast} (X; \mathbb{Q}) \to H^{\ast}(\tilde{X}, \mathbb{Q})$ is injective, which allows to reduce the argument to the case of a trivial action. QED

After I have written all this, I realize that the main point is that the bundle $E$ has structural group $O(p,q)$ and it is flat as such a bundle. It allows a reduction of the structural group to $O(p) \times O(q)$ (by the contractibility of $Q(b)$), but not as a flat bundle. If you find a flat reduction, then $ch(Ind(D))=0$ (in positive degrees) by Chern-Weil theory. The index theorem is not really relevant here; it identifies $ch(ind(D))$ with the fibre integral of the $L$-class of the vertical tangent bundle.