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For example given a tringle with each side equal to $1$ and angles between each side being perpendicular, then this triangle is on a quarter of sphere with radius $\dfrac {2}{\pi}$.

Is this example of triangle in elliptic geometry? what would example of equilateral triangle in hyperbolic geometry look like?

Is there a general formula to determine the surface type that tringle is bounding ?

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    Why, that's simple: if the sum of angles exceeds $180^\circ$, it is elliptic geometry, otherwise it is either planar ($sum=180^\circ$) or hyperbolic ($sum<180^\circ$). – 2017-01-22
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    See the Girard's theorem for the area of a spherical triangle (http://theoremoftheday.org/GeometryAndTrigonometry/Girard/TotDGirard.pdf) which depend **only** on the angles of the triangle. – 2017-01-22

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The problem is that your question has no answer, but not for the reason you might expect: take a 30-60-90 triangle with sides 1, 2, $\sqrt{3}$. Then "the surface type that the triangle is bounding" is the plane. But it's also the cylinder, or the "flat torus" in 4-space, and indeed it could be part of a two-holed-torus (albeit one that didn't "look" like the usual one") or of a projective plane in 5-space, or ...

Now if you've got a particular collection of surface families that you're interested in -- spheres, the plane, perhaps some others -- then there might be a unique answer.

But what you can say is the following:

If the triangle edges actually bound a topological disk on the surface, and if the edges are all geodesics on the surface, then the total curvature of the region bounded can be computed from the Gauss-Bonnet theorem, and you may find that the total curvature is either positive, negative, or zero. If you're restricting your attention to the case of surfaces of constant curvature, then this will indeed tell you the "type" of surface.

Note that the restrictions in the first sentence are necessary: you can draw a triangle (three vertices connected by three edges) that goes "around the handle" of a coffee cup, and this triangle does not bound a disk on the surface of the coffee cup, so you can't say anything about "the surface type the triangle is bounding". And it's also essential that the edges of the triangle be geodesics on that surface, or the mere edge-lengths and angles don't suffice to determine the total curvature of the bounded region -- you need also to know the linear curvature along each edge.