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2 Added a proof of my claim.

When you have a vector bundle on a manifold $X$ with boundary, trivialised over $\partial X$, there are characteristic classes valued in $H^\ast (X,\partial X)$. Here, when $L$ is orientable, the Maslov index is twice the first Chern class of $u^\ast TM$ relative to the trivialisation on the boundary induced by $L$, evaluated on $[\Sigma,\partial \Sigma]$.

When $\Sigma$ is closed, the Chern number of $u^\ast TM$ is the signed count of zeroes of a transversely-vanishing section $s$ of $u^\ast\Lambda^{max}_{\mathbb{C}}TM$.

When there is an orientable boundary condition, the relative Chern number is the same thing, but you choose $s$ non-vanishing along the boundary and tangent to the real line sub-bundle $\Lambda^{max}_{\mathbb{R}} TL$.

This doesn't make sense when $u^*|_{\partial \Sigma} TL \to \partial \Sigma$ is not orientable: its top exterior power then has no non-vanishing section. Besides, the Maslov index is odd in this case.

ADDED: Here's a proof using the method of Robbin's appendix to McDuff-Salamon ("$J$-holomorphic curves and symplectic topology"). Robbin characterises the boundary Maslov index as an invariant of bundle pairs (complex vector $E$ bundle over a surface, totally real sub-bundle $F$ over the boundary) which is additive under direct sum and under sewing boundaries and is suitably normalised for line bundles over the disc. The uniqueness proof, by "pair-of pants induction", still applies when $F$ is assumed orientable. The invariant "twice the relative Chern number" evidently satisfies the direct sum and sewing properties, and the section $z\mapsto z$ of the trivial line bundle over the disc satisfies the standard Maslov-index 2 boundary condition. Done!

1

When you have a vector bundle on a manifold $X$ with boundary, trivialised over $\partial X$, there are characteristic classes valued in $H^\ast (X,\partial X)$. Here, when $L$ is orientable, the Maslov index is twice the first Chern class of $u^\ast TM$ relative to the trivialisation on the boundary induced by $L$, evaluated on $[\Sigma,\partial \Sigma]$.

When $\Sigma$ is closed, the Chern number of $u^\ast TM$ is the signed count of zeroes of a transversely-vanishing section $s$ of $u^\ast\Lambda^{max}_{\mathbb{C}}TM$.

When there is an orientable boundary condition, the relative Chern number is the same thing, but you choose $s$ non-vanishing along the boundary and tangent to the real line sub-bundle $\Lambda^{max}_{\mathbb{R}} TL$.

This doesn't make sense when $u^*|_{\partial \Sigma} TL \to \partial \Sigma$ is not orientable: its top exterior power then has no non-vanishing section. Besides, the Maslov index is odd in this case.