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Let $f\colon X\to Y$ be a smooth surjective map of smooth manifolds of dimension $n$ which are not necessarily orientable. A relative orientation of $X$ over $Y$ consists of an isomorphism $\psi\colon L\otimes L\to\textrm{Hom}(f^*\Omega_Y^n,\Omega_X^n)$ where $L$ is a line bundle on $X$. For every $y\in Y$ outside the branch locus of $f$, for a small enough neighbourhood $U\subseteq Y$ of $y$ and an orientation of $U$, such a relative orientation determines an orientation of $f^{-1}(U)$. Can we define this solely in topological terms, not making use of the smooth structure?

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Yes, if I understand the question right.

Reading between the lines, I suppose that you would define an orientation of a smooth $n$-manifold $X$ to be an isomorphism $L\otimes L\to \Omega^n_X$ where $L$ is a line bundle on $X$. That seems a little funny to me. I might have defined an orientation as a trivialization of the line bundle $\Omega^n_X$, or rather an equivalence class of such trivializations with respect to multiplying by positive functions. Of course, it's true that a line bundle is trivial if and only if it is isomorphic to $L\otimes L$ for some $L$; but are we really interested in isomorphisms $L\otimes L\to \Omega^n_X$ for nontrivial $L$? Any trivialization of $\Omega^n_X$ that we can make with a nontrivial $L$ we can also make with a trivial $L$.

Anyway, a topological $n$-manifold has a canonical line bundle on it, whose fiber at $x\in X$ is the relative cohomology group $H^n(X,X\setminus x;\mathbb R)$ (or you can use homology). Let's call it $L_X$. This is isomorphic to $\Omega^n_X$. You can define an orientation of $X$ to be a trivialization of $L_X$, and define a relative orientation to be a trivialization of $\operatorname{Hom}(f^\ast L_Y,L_X)$ (in both cases up to multiplying by positive functions).

Or if you don't want to have to mention those positive scaling factors you can use $H^n(X,X\backslash x;\mathbb Z)$ instead, getting a bundle of infinite cyclic groups rather than a bundle of one-dimensional vector spaces.

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