I'm getting through a bit of confusion... Consider a plane. Give it hyperbolic geometry. If embedded in a euclidean tridimensional space, will it look "saddle-shaped"? And what about an embedding in a hyperbolic space? Will it look flat? However, if a hyperbolic plane has a defined (negative) Gauss curvature - and that's an intrinsic feature - doesn't it mean that such feature will be the same regardless to the space of embedding?
To a differential geometer, there is a difference between intrinsic geometry and extrinsic geometry.
The intrinsic geometry of a hyperbolic plane, for example the fact that it has constant Gauss curvature equal to $-1$, is independent of its extrinsic features under various different embeddings. Your question gives two excellent examples of this phenomenon: as you suggest, any isometric embedding of any little piece of the hyperbolic plane into flat space must indeed have an extrinsic saddle shaped appearance; and there is indeed an isometric embedding of the hyperbolic plane into hyperbolic space that has a flat extrinsic appearance.
But this is not a contradiction.
In fact, there are theorems in differential geometry which I won't try to state precisely, but I'll summarize intuitively. Namely, for any embedding of the hyperbolic plane $P$ into a higher dimensional Riemannian manifold $M$, and for any $x \in P$, there is a formula relating three features to each other: the intrinsic geometry at $x$ of $P$; the extrinsic geometry at $x$ of the embedding of $P$ into $M$; and the intrinsic geometry of $M$ at $x$.
So no, there is nothing to enforce a constant extrinsic appearance of an isometric embedding of the hyperbolic plane into a higher dimensional manifold.