Here's proof for conclusion 1; I think 2 will then follow fairly easily. For brevity, I'll write things like $p\upharpoonright\eta$ when I really mean its extension by 1 to domain $\nu$. Suppose $\eta$ were a counterexample to 1. Since $p\upharpoonright\eta$ fails to force $\eta\leq\dot\alpha$, it must have an extension $q$ forcing $\dot\alpha$ to have some specific value $\xi<\eta$. Apply promptness to infer that $q\upharpoonright\xi$ already forces this value $\xi$ for $\dot\alpha$. But, since $\xi<\eta$ and $q$ extends $p\upharpoonright\eta$, the conditions $q\upharpoonright\xi$ and $p$ are compatible. Since the former forces $\dot\alpha$ to have value $\xi$ while the latter forces $\eta\leq\dot\alpha$, this is a contradiction.

EDIT: I think I was too optimistic in expecting conclusion 2 to "follow fairly easily", so I'm adding information about that. Notice first that what I wrote above remains correct if we replace the inequalities $\eta\leq\dot\alpha$ and $\xi<\eta$ by $\eta<\dot\alpha$ and $\xi\leq\eta$, respectively. (Of course, $\eta<\dot\alpha$ is equivalent to $\eta+1\leq\dot\alpha$, so I could apply the preceding paragraph directly, but then I'd get a conclusion about $p\upharpoonright(\eta+1)$, whereas I really want $p\upharpoonright\eta$.)

Next, let me establish a "dual" (i.e., order-reversed) version of conclusion 1, namely that if $\dot\alpha$ is prompt and $p$ forces $\dot\alpha\leq\eta$, then $p\upharpoonright\eta$ already forces the same. To prove this, suppose it fails, and let $q$ be an extension of $p\upharpoonright\eta$ forcing $\eta<\dot\alpha$. By what I proved above, $q\upharpoonright\eta$ suffices to force $\eta<\dot\alpha$. But $q\upharpoonright\eta$ is an extension of $p\upharpoonright\eta$ and is therefore compatible with $p$. That's absurd since compatible conditions can't force contradictory things about the ordering of $\eta$ and $\dot\alpha$.

Of course, this dual version of conclusion 1 also has an analog for conditions $p$ that force $\dot\alpha<\eta$.

At last, I'm ready to prove the part of conclusion 2 that deals with the supremum, say $\dot\beta$, of some prompt names $\dot\alpha_i$. So suppose, toward a contradiction, that $p$ forces $\dot\beta=\xi$ but $p\upharpoonright\xi$ doesn't force this. There are two possibilities: Either some extension of $p\upharpoonright\xi$ forces some $\dot\alpha_i$ to be strictly above $\xi$, or some extension of $p\upharpoonright\xi$ forces some ordinal $\eta<\xi$ to be an upper bound for all the $\dot\alpha_i$'s. In either case, let $q$ be such an extension of $p\upharpoonright\xi$.

Consider the first case: $q$ forces $\xi<\dot\alpha_i$ for a certain index $i$. Then, by one of the versions of conclusion 1, $q\upharpoonright\xi$ already forces the same inequality. But $q\upharpoonright\xi$ is an extension of $p\upharpoonright\xi$ and is therefore compatible with $p$, which forces the opposite. So this case cannot occur.

There remains the case that, for a certain $\eta<\xi$, the condition $q$ forces $\dot\alpha_i\leq\eta$ for every index $i$. By a dual version of conclusion 1, as proved above, $q\upharpoonright\eta$ forces all these inequalities and therefore forces $\dot\beta\leq\eta$. But $q\upharpoonright\eta$ is compatible with $p$, which forces the opposite. So this case is also impossible, and the proof of conclusion 2 for the supremum is complete.

Finally, since I was (I hope) careful to use, in this proof for the supremum, only versions of conclusion 1 for which the dual is also available, we can dualize this whole proof to get the result for the infimum, which is of course the minimum since we're dealing with a well-rodering.