I would answer question 1 by saying that
$M=\text{Ult}(V,U)$ if and only if $j$ is isomorphic to an
ultrapower by *some* normal measure. This is another way of
saying that normal measures are minimal with respect to the
Rudin-Kiesler order. The point is that an embedding is the
ultrapower by a normal measure if and only if it is the
ultrapower by its induced normal measure, if and only if
$\kappa$ as a seed generates the whole embedding, in the
sense that every element of $M$ has the form
$j(f)(\kappa)$.

For question 2, suppose that $j:V\to M$ witnesses the
$\lambda$-supercompactness of $\kappa$, so that $\kappa$ is
the critical point of $j$ and $M^\lambda\subset M$. Your
ultrafilter $U$ is what is sometimes called the *induced
normal measure* of $j$, and since the seed hull
$X_\kappa=\{j(f)(\kappa)\mid f:\kappa\to V\}$ is an
elementary substructure of $M$, we may collapse it and form
a commutative diagram, with $V\to M_U\to M$, where
$j_U:V\to M_U=\text{Ult}(V,U)$ is the ultrapower by $U$ and
$k:M_U\to M$ is the inverse collapse of $X_\kappa$ and
$j:V\to M$ is the composition $j=k\circ j_U$.

*insert triangular commutative diagram here*

Since $M^\lambda\subset M$ in $V$ and $M_U\subset V$, it
follows immediately that $M^\lambda\cap M_U\subset M$.
Thus, $M$ remains $\lambda$-closed for sequences in $M_U$.
However, when $\lambda\gt\kappa$, then it will never be the
case that $M\subset M_U$. This is because $M_U$ is an
ultrapower by a filter on $\kappa$ and therefore $j_U$ is
continuous at ordinals of cofinality $\kappa^+$. Therefore,
$j_U(\kappa^+)$ has true cofinality $\kappa^+$ in $V$ and
hence also in $M$, but it is regular in $M_U$, being a
successor cardinal there. Thus, although the quotient map
$k:M_U\to M$ is a $\lambda$-closed embedding, it can never
be an internal embedding from the perspective of $M_U$,
since $M$ is not a subclass of $M_U$.

However, if you are willing to consider the case
$\lambda=\kappa$, then it is possible for $k:M_U\to M$ to
be internal to $M_U$, for we may simply let $j$ be the
ultrapower by $U\times U_1$ for any other measure on
$\kappa$, so that $k$ is simply the ultrapower in $M_U$ by
$j_U(U_1)$. In this case, $M$ is $j_U(\kappa)$-closed in
$M_U$, since it is an internal ultrapower there by a
measure on $j_U(\kappa)$.