Existence of an isometric embedding into Euclidean space with bounded second fundamental form

The curvature tensor can be expressed in terms of second fundamental form. Therefore bounded curvature is a necessary condition.

Yet injectivity radius has to be bounded below.

These two conditions are sufficient. It seems that this could be proved along the same lines as the Nash embedding theorem.

P.S. In the formulation, you had to say what you mean by "second fundamental form". Most people think it is only defined for hypersurfaces, but you mean a quadratic form on tangent space with values in the normal space.

The Gauss Equations are able to give you some coarse information immediately, see for instance the wikipedia article http://en.wikipedia.org/wiki/Gauss–Codazzi_equations here. For instance, if $M$ is $n$ dimensional, and you have such an isometric embedding, then about some point $p \in M\cap \mathbb{R}^N$ there is a basis of $N-n$ vectorfields, say $\{e_i\}$ normal to $M$ in $R^N$. Then for each $e_i$ one gets an operator $X \to \nabla_X e_i$ so in this sense there are $N- n$ second fundamental forms. Now the gauss equation writes the curvature of $M$ in terms of the sum of these operators, so if $M$ has unbounded curvature, so must this sum. In particular if $N = n + 1$ then a necessary condition for an embedding (with bounded second fundamental form) is that the curvature of $M$ is bounded.