As some of the comments suggest, "semisimple" correlates well for the groups and their Lie algebras in characteristic 0, but "reductive" doesn't work so well. For a Lie algebra in characterisic 0, Bourbaki Groupes et algebres de Lie (Chapter I, section 6.4) defines it to be reductive just when its adjoint representation is completely reducible; other convenient equivalent conditions follow. The groups don't show up until later chapters, where semisimple Lie groups are emphasized.
In your situation, $G$ can equally well be viewed as a semisimple algebraic group (from the viewpoint of Chevalley's classification over arbitrary algebraically closed fields). On the other hand, the Zariski-closed subgroup $H$ is also closed in the euclidean topology and thus is itself a Lie subgroup of $G$. But for example $H$ might well be commutative and unipotent (as an algebraic group), hence connected in either topology; then its Lie algebra is abelian and thus "reductive" even though $H$ itself isn't. (You might as well assume $H$ is connected in any case, since a finite component group doesn't affect linear reductivity in characteristic 0.)
Concerning the notion of "linearly reductive", it has been knwn since early work of Nagata and others that this is equivalent for algebraic groups to being reductive in charactristic 0: almost-direct product of central torus and semisimple derived group. But unlike the notion of "semisimple", there is no automatic equivalence between an algebraic group being reductive and its Lie algebra being reductive. (Most of the literature is old, including Benjamin lecture notes by John Fogarty. The 1975 Annals of Math. paper by Haboush has references, but worked in the prime characteristic setting to prove Mumford's conjecture that all reductive groups are "geometrically reductive".)