In chapter 3 of "Characteristic Classes", Milnor defines bundle maps, requiring them to map fibers isomorphically onto fibers. Why not merely require homorphisms on fibers? (e.g., for the given definition, the inclusion map of a subbundle doesn't qualify as a bundle map.) Is it a matter of convenience, or is there something else going on?
Thanks!
One can define characteristic classes in a more general context. For any topological group $G$, there is a universal principal $G$-bundle (henceforth just $G$-bundle) $EG \to BG$. $BG$ is called the classifying space of $G$, because isomorphism classes of principle $G$-bundles over $X$ correspond bijectively to homotopy classes of maps $f: X \to BG$, the correspondence taking a representative map, and defining the pullback bundle $f^*(EG \to BG)$, a $G$-bundle on $X$. (This turns out to depend only on the homotopy class of $f$.)
Now, pick an $\alpha \in H^*(BG;A)$ for some commutative ring $A$. We define the "$\alpha$-characteristic class" (this is not standard terminology - I just made it up) for a $G$-bundle $E \to X$ to be the class $f^*\alpha \in H^*(X;A)$, where $f: X \to BG$ is the defining map for our $G$-bundle. Then if $g: Y \to X$ is a continuous map, the $\alpha$-characteristic class of the pullback bundle $g^*(E \to X)$ is precisely $g^*(f^*(\alpha))$. You might notice this is the way Milnor's characteristic classes transform...
Indeed, to any real $n$-dimensional vector bundle, there is a canonically assigned $O(n)$-bundle (and this correspondence is again bijective on isomorphism classes). One may calculate (indeed, Milnor-Stasheff do) that $H^*(BO(n),\Bbb Z/2\Bbb Z) \cong \Bbb Z_2[w_1, w_2, \dots, w_n]$, a polynomial algebra with $|w_i| = i$. For any real, $n$-dimensional vector bundle $E \to X$, let $f$ be the representing map $X \to BO(n)$; then $w_i(E) = f^*(w_i)$. One may do something analagous for complex vector bundles ($\cong$ $U(n)$-bundles); this gives Chern classes.
Now, what Milnor and Stasheff call a bundle map $E \to E'$, where $E \to X$ is a vector bundle and so is $E' \to Y$, is precisely the same as a map $f: X \to Y$ such that $f^*(E') = E$. So the above discussion makes it clear this is how characteristic classes transform: under pullback maps, because that's precisely how we get a map $H^*(Y;A) \to H^*(X;A)$ in cohomology!
Indeed, if you tried to define how characteristic classes transform by some '$f^*$' under your sort of bundle map, we no longer have the formula $f^*(w_i(E')) = w_i(E)$; for take $E$ to be the trivial 0-dimensional bundle over the 1-point space, and $E'$ to be literally any vector bundle; the formula would say that $w_i(E') = 0$ for all $E'$. If you try to weaken your hypotheses a little bit (maybe only homomorphisms between $n$-dimensional vector bundles?) you get the same problem.