One of the basic facts of product forcing is the following,
which appears in any of the standard accounts of forcing:

**Theorem.** If $M$ is a model of ZFC and
$\mathbb{P},\mathbb{Q}$ are forcing notions in $M$, with
$M$-generic filters $G\subset\mathbb{P}$ and
$H\subset\mathbb{Q}$, then the following are equivalent:

$G$ is $M[H]$-generic for $\mathbb{P}$.

$H$ is $M[G]$-generic for $\mathbb{Q}$.

$G\times H$ is $M$-generic for the product forcing $\mathbb{P}\times\mathbb{Q}$.

And this is what it means for $G$ and $H$ to be *mutually*
$M$-generic.

Thus, the first answer to your question is that there are
mutually $M$-generic filters for $\mathbb{P}$ and
$\mathbb{Q}$ precisely when there are any $M$-generic
filters for the product forcing. For example, if $M$ is a
countable model, then these always exist.

The following observation of Solovay gives one of the most
important relations between $M[G]$ and $M[H]$.

**Theorem.** If $G$ and $H$ are mutually $M$-generic,
then $M[G]\cap M[H]=M$.

Proof. Suppose that $x\in M[G]\cap M[H]$. Then
$x=\tau_G=\sigma_H$ for $\mathbb{P}$-name $\tau$ and
$\mathbb{Q}$-name $\sigma$. We may assume by
$\in$-induction that $x\subset M$. Because of the equality,
there must be a condition $(p,q)\in G\times H$ forcing that
$\tau=\sigma$, and we may also assume $p$ and $q$ force
$\tau\subset\check M$ and $\sigma\subset\check M$. (Note,
we may canonically regard $\mathbb{P}$ and
$\mathbb{Q}$-names as $\mathbb{P}\times\mathbb{Q}$-names.)
It now follows, however, that $p$ decides $\check y\in
\tau$ for every $y\in M$, for if not, then then there would
be stronger conditions, some forcing some $\check y\in\tau$ and
others forcing $\check y\notin \tau$. Let $p'$ be stronger
than $p$, forcing an opposite answer to whether
$y\in\sigma_H$, forced by some $q'\in H$. Thus, $(p',q')$
is stronger than $(p,q)$, and forces $\tau\neq\sigma$, a
contradiction. QED

Finally, let me also mention an interesting fact about
mutual genericity, which I heard years ago from Woodin.

**Theorem.** If $M$ is a countable model of ZFC, then there are $M$-generic Cohen reals
$c$ and $d$ such that $M[c]$ and $M[d]$ have no common
forcing extension. Thus, the models $M[c]$ and $M[d]$ are non-amalgamable,
which makes them very far from mutually generic.

Proof. Fix a real $z$ that cannot possibly be
added by forcing over $M$, such as a real coding all of
$M$. Now, enumerate the dense subsets $D_n$ of Cohen
forcing in $M$, and build $c$ and $d$ in stages, $c=\cup_n
c_n$ and $d=\cup_n d_n$. First, let $c_0$ be a finite
binary sequence in the first dense set, and let $d_0$ be
all zeros to the same length, then $1$, then the first bit
of $z$, and then extend to an element of $D_0$. Now extend
$c_0$ to $c_1$ with all $0$s to the length of $d_0$, then
$1$, then get into $D_1$, and so on. Each of $c$ or $d$
individually will be $M$-generic, but any model containing
both of them will be able to define $z$. So $M[c]$ and
$M[d]$ cannot be contained in any model $N\models$ZFC with
the same ordinals. QED

One can easily arrange more complicated patterns, with
three reals, for example, any two of which are mutually
generic, but such that the three of them together are
non-amalgamable.

Meanwhile, it is a theorem of mine (joint with G. Fuchs and J. Reitz)
that if you have a
countable family of forcing extensions $M[G_n]$ of a
countable model $M$, whose forcing notions have uniformly
bounded size in $M$, such that any finite subfamily is
amalgamable, then there is a common forcing extension
$M[H]$ with $M[G_n]\subset M[H]$ for every $n$. You can find the result in
our recent paper on set-theoretic geology.