You need some additional assumptions to get this (just as you need assumptions on I just to get that the ultrapower is well-founded).
In Matt Foreman's chapter for the Handbook of Set Theory, he states the following theorem (it is Theorem 2.25 in the preliminary version I have here, but the published number may differ).
Theorem. Suppose I is a normal, fine, precipitous ideal on $Z\subset P(X)$, where $|X|=\lambda$. Let $G\subset P(Z)/I$ be generic, and $M$ the generic ultrapower of $V$ by $G$. Then $P(\lambda)\cap V\subset M$. Further, if $I$ has the disjointing property, then $M^\lambda\cap V[G]\subset M$.
Note that this theorem covers your case of $Z=P_\kappa(\lambda)$.
To prove the first part, you simply observe that $[id]$ represents $j " \lambda$, and then for any $A\subset\lambda$ you can get $j"A$ using the function $g(z)=z\cap A$. Now, from $j"\lambda$ and $j"A$ you can easily build $A$ in $M$.
For the second part, the part you were interested in, you use the disjointing property in order to know that a term for a $\lambda$-sequence of elements of $M$ can be transformed into a $\lambda$-sequence of terms in $M$. That is, if $\langle\dot a_\alpha :\alpha<\lambda\rangle$ is a $\lambda$-sequence of terms for objects in $M$, then disjointing allows us to find in $V$ a sequence of functions $\vec g = \langle g_\alpha: \alpha<\lambda\rangle$ such that $[g_\alpha]^G = \dot a_\alpha^G$. From this, it follows that the function $g(z) = \langle g_\alpha(z) | \alpha\in z\rangle$ represents $j(\vec g)(j"\lambda)$, which is $\langle j(g_\alpha)_\beta(j"\lambda) | \beta\in j"\lambda\rangle$, from which we can construct $\langle j(g_\alpha)(j"\lambda) | \alpha <\lambda\rangle$, which is the desired $\lambda$-sequence.