What you get is that if j:V to M is the ultrapower by any ultrafilter U on any set X, then every element of M has the form j(f)([id]). You can prove this by building an isomorphism from the ultrapower to the sets of this form. This way of thinking is also known as "seed theory".

**Theorem.** Suppose that j:V to M is an elementary embedding of the universe V into M. Then j is the ultrapower map by a measure on a set if and only if there is some s in M such that every element of M has the form j(f)(s).

That is, the ultrapower embeddings are precisely the embeddings whose target is generated by a single element.

Proof. If j is the ultrapower by U on X, then let s=[id], and argue that [f]_U is j(f)(s). Conversely, if M = { j(f)(s) | f in V }, where we assume that f is a function on some set X such that a ∈ j(X), then define the measure U by A ∈ U iff s ∈ j(A). This is a κ-complete ultrafilter on P(X). One can show that Ult(V,U) is isomorphic to M, by mapping [f]_U to j(f)(s). QED

**Theorem.** If U is a κ complete ultrafilter on κ with ultrapower embedding j:V to M, then every element of M has the form j(f)(κ) if and only if U is isomorphic to a normal measure.

Proof. You know the backwards implication. For the forward implication, suppose that every element of M has form j(f)(κ). In particular, β = j(f)(κ), where β = [id]_U is the seed for U. But also, κ = j(g)(β), where g(α) is the smallest ξ for which f(ξ)=α. Let μ = { X subset κ | κ ∈ j(X) } be the induced normal measure. Note that X in μ iff j(g)(β) in j(X) iff β in j(g^{-1}X) iff g^{-1}X in U. So μ is Rudin-Kiesler below U. Also, U is Rudin-Kiesler below μ since X in U iff f^{-1}X in μ. So μ and U are isomorphic.QED

One may illustrate the situation with product measures. Suppose that U is normal. The product measure UxU is isomorphic to the two-step iteration, where j_0:V to M is the ultrapower by U, and h:M to N is the ultrapower in M by j_0(U). Every element of M has the form j_0(f)(κ), and every element of N has the form h(g)(κ_{1}), where κ_{1} = j_0(κ). If j is the composition of j_0 and h, then j:V to N and every element of N has the form j(f)(κ, κ_{1}). If one only looks at j(f)(κ) inside N, then you will only get ran(h), which is isomorphic to M, but not all of N. So this would be a counterexample to what you asked about.