Suppose $M$ is a transitive class and $j : V\to M$ is an elementary embedding.

Some notation.

• For any $x\in M$, let $H_x = \{j(f)(x) : f\text{ is a function}\}\prec M.$
• Let $j_x : V\to M_x$ be the ultrapower by the derived ultrafilter $U_x = \{A\subseteq X : x\in j(A)\}$ where $X$ is any set such that $x\in j(X)$.
• Let $k_x : M_x\to M$ be the unique elementary embedding with $k_x\circ j_x = j$ and $k_x([\text{id}]_{U_x}) = x$.

Equivalently, $M_x$ is the transitive collapse of $H_x$, $k_x : M_x\to M$ is the inverse of the transitive collapse, and $j_x = k_x^{-1}\circ j$.

A nonstandard definition. An ordinal $\xi$ is a weak generator of $j$ if $\xi\neq j(f)(\alpha)$ for any function $f$ and any $\alpha < \xi$. (The difference between a generator and a weak generator is that in the definition of a generator one requires that $\xi\neq j(f)(\vec \alpha)$ for any $\vec \alpha \in [\xi]^{<\omega}$. Thus a generator is the same thing as an additively indecomposable weak generator.)

A useful fact about elementary embeddings. Suppose $x\in M$ and $\xi$ is the least ordinal such that $x \in H_\xi$. Then $$H_x = H_\xi$$ Thus $j_x = j_\xi$ and $k_x = k_\xi$. Moreover $\xi$ is a weak generator of $j$. We call $\xi$ the weak generator associated to $x$.

Key point: In your situation, for any distinct weak generators $\xi,\xi'$ of $j$, $j_\xi\neq j_{\xi'}$. (Also weak generators are the same thing as generators in your situation, but this is not really relevant.) Since $V_{\kappa+2}\subseteq M$, if $\xi < (2^\kappa)^+$ is a weak generator of $j$, then $$\xi < ((2^{\kappa})^+)^{M_\xi} = \min (\text{Ord}\setminus H_\xi)$$ Hence $((2^{\kappa})^+)^{M_\xi}$ is the least weak generator of $j$ above $\xi$. Therefore for distinct weak generators $\xi,\xi'$, $((2^{\kappa})^+)^{M_\xi}\neq ((2^{\kappa})^+)^{M_{\xi'}}$, so $M_\xi \neq M_{\xi'}$, and certainly $j_\xi \neq j_{\xi'}$.

The answer. Assume that for any distinct weak generators $\xi,\xi'$ of $j$, $j_\xi\neq j_{\xi'}$. Suppose $X$ is a set, suppose $x,x'\in j(X)$, and assume the derived ultrafilter $U_x = \{A\subseteq X : x\in j(A)\}$ is equal to $U_{x'} = \{A\subseteq X : x'\in j(A)\}$. We will show $x = x'$. Let $\xi$ and $\xi'$ be the weak generators associated to $x$ and $x'$ respectively. Since $U_x = U_{x'}$, the associated ultrapower embeddings $j_x$ and $j_{x'}$ are also equal. It follows that $j_\xi = j_x = j_{x'} = j_{\xi'}$. Thus by our assumption about weak generators, $\xi = \xi'$. In particular, $k_x = k_{\xi} = k_{\xi'} = k_{x'}$. Therefore $$x = k_x([\text{id}]_{U_x}) = k_{x'}([\text{id}]_{U_{x'}}) = x'$$ Hence $x = x'$, as desired.

TLDR: Yes, all the ultrafilters are different.

Suppose $M$ is a transitive class and $j : V\to M$ is an elementary embedding.

Some notation.

• For any $x\in M$, let $H_x = \{j(f)(x) : f\text{ is a function}\}\prec M.$
• Let $j_x : V\to M_x$ be the ultrapower by the derived ultrafilter $U_x = \{A\subseteq X : x\in j(A)\}$ where $X$ is any set such that $x\in j(X)$.
• Let $k_x : M_x\to M$ be the unique elementary embedding with $k_x\circ j_x = j$ and $k_x([\text{id}]_{U_x}) = x$.

Equivalently, $M_x$ is the transitive collapse of $H_x$, $k_x : M_x\to M$ is the inverse of the transitive collapse, and $j_x = k_x^{-1}\circ j$.

A nonstandard definition. An ordinal $\xi$ is a weak generator of $j$ if $\xi\neq j(f)(\alpha)$ for any function $f$ and any $\alpha < \xi$. (The difference between a generator and a weak generator is that in the definition of a generator one requires that $\xi\neq j(f)(\vec \alpha)$ for any $\vec \alpha \in [\xi]^{<\omega}$. Thus a generator is the same thing as an additively indecomposable weak generator.)

A useful fact about elementary embeddings. Suppose $x\in M$ and $\xi$ is the least ordinal such that $x \in H_\xi$. Then $$H_x = H_\xi$$ Thus $j_x = j_\xi$ and $k_x = k_\xi$. Moreover $\xi$ is a weak generator of $j$. We call $\xi$ the weak generator associated to $x$.

Key point: In your situation, for any distinct weak generators $\xi,\xi'$ of $j$, $j_\xi\neq j_{\xi'}$. (Also weak generators are the same thing as generators in your situation, but this is not really relevant.) Since $V_{\kappa+2}\subseteq M$, if $\xi < (2^\kappa)^+$ is a weak generator of $j$, then $$\xi < ((2^{\kappa})^+)^{M_\xi} = \min (\text{Ord}\setminus H_\xi)$$ Hence $((2^{\kappa})^+)^{M_\xi}$ is the least weak generator of $j$ above $\xi$. Therefore for distinct weak generators $\xi,\xi'$, $((2^{\kappa})^+)^{M_\xi}\neq ((2^{\kappa})^+)^{M_{\xi'}}$, so $M_\xi \neq M_{\xi'}$, and certainly $j_\xi \neq j_{\xi'}$.

The answer. Assume that for any distinct weak generators $\xi,\xi'$ of $j$, $j_\xi\neq j_{\xi'}$. Suppose $X$ is a set, suppose $x,x'\in j(X)$, and assume the derived ultrafilter $U_x = \{A\subseteq X : x\in j(A)\}$ is equal to $U_{x'} = \{A\subseteq X : x'\in j(A)\}$. We will show $x = x'$. Let $\xi$ and $\xi'$ be the weak generators associated to $x$ and $x'$ respectively. Since $U_x = U_{x'}$, the associated ultrapower embeddings $j_x$ and $j_{x'}$ are also equal. It follows that $j_\xi = j_x = j_{x'} = j_{\xi'}$. Thus by our assumption about weak generators, $\xi = \xi'$. In particular, $k_x = k_{\xi} = k_{\xi'} = k_{x'}$. Therefore $$x = k_x([\text{id}]_{U_x}) = k_{x'}([\text{id}]_{U_{x'}}) = x'$$ Hence $x = x'$, as desired.