If my understanding of $\mathbb{P}$-names $\tau$ is correct, they must be comprehensible by someone living in the ground model $\mathcal{M}$.
However, the value $\tau[G]$ of $\tau$ need not be intelligible in the ground model $\mathcal{M}$ since one requires knowledge of the generic set $G$ in order to understand $\tau[G]$, but such knowledge is not available in $\mathcal{M}$ as $G \not\in \mathcal{M}$ in general.
Suppose that what I just said is true (please correct me if not), I struggle to see why $\tau$ is comprehensible in $\mathcal{M}$.
Here is my definition of $\mathbb{P}$-names (again, correct me if I'm wrong) :
Let $\mathbb{P}$ be a forcing notion of $\mathcal{M}$ and let $\alpha \in \text{Ord}$. Define recursively the hierarchy of $\mathbb{P}$-names $\{N_\alpha :~~ \alpha \in \text{Ord}\}$ as follows : $\tau \in N_\alpha$ if and only if every element of $\tau$ is an ordered pair $\langle \sigma, p\rangle$ where $\sigma \in N_{<\alpha}$ and $p \in P$. That is to say $$\tau ~\subseteq~ N_{<\alpha} \times P ~=~ \bigcup\limits_{\beta < \alpha} N_\beta \times P$$
Yes, a name is something which exists in the ground model.
Look at the full binary tree. Given any node on the tree, you can't quite say what is the entire real number that it will define, but you can say that the node is limiting your options for which real number you will end up with.
Forcing is about the same. As you "progress along" the forcing poset you gain more information, but you can't quite know what it will entail. But you can know that once you've gone pass a certain condition in the forcing, you will have a certain information regarding the generic filter, or any other name.
And this is the essence of it. You can say, from $\cal M$, which condition forces what, but you can't quite "pack it up" into a single object within $\cal M$. For this you need the generic filter.