Ok I agree. Thurston gave a nice example of how the normal bundle can have unbounded curvature (but not if $N = n+1$) and and Deane mentioned that this can't happen if the second fundamental form is bounded. So my answer as stated is correct, and you have answered your own question in your first comment :-)

You are right, in that sense you can talk about "the" second fundamental form, but then a bound on that operator is the same as individually bounding each of the operators associated to the $e_i$ that I mentioned above.

As another side note, if you don't care how large N is, then, if M has bounded curvature, you can use the nash embedding theorem to embed M in to R^N, and then the gauss equations will tell you that each second fundamental form is bounded. The really hard part about isometric embeddings is bounding N.

If $N > n+1$ then there isn't "the" second fundamental form. You can think about the "bending" of $M$ in $R^N$ in any direction orthogonal to $M$. In this case, the bounded curvature of $M$ is still a necessary condition if you require that all of the "second fundamental forms" are bounded. If you let some of them go to infinity, then you can let some of the curvatures of $M$ go to infinity.