Problem understanding a passage of the proof of $\mathfrak{p}=\mathfrak{t}$ involving forcing
I think the problem is slightly more basic. Fremlin has the fully correct $$ D \Vdash \check D\in\dot{\mathcal{G}} $$ Also, Fremlin did not fix one generic $G$ at the outset; he works with names and the forcing relation everywhere.
Suppose $\alpha < \beta$ and yet $Y_\beta \not \subseteq_* Y_\alpha$. Let $B'$ be the infinite set of all $n \in B$ such that $Y_{\beta, n} := Y_\beta \cap (\{n\} \times \omega^{<\omega}) \not \subseteq Y_{\alpha, n} := Y_\alpha \cap (\{n\} \times \omega^{<\omega})$. For each $n \in B'$, choose $t_n \in Y_{\beta, n} \backslash Y_{\alpha, n}$. Then $t_n$ extends $f_\beta(n)$ but does not extend $f_\alpha(n)$, so $f_\alpha(n) \not \trianglelefteq f_\beta(n)$, so $B'$ forces that $f_\alpha/\dot{G} \not \trianglelefteq f_\beta/\dot{G}$.
This is more intuitive if we reason inside of $\mathbb{V}^\omega/\dot{G}$: let $\hat{\omega}$ denote the $\omega$ of $\mathbb{V}^\omega/\dot{G}$ (this is a $Q$-name). For each $\alpha < \theta$, write $\hat{s}_\alpha = f_\alpha/\dot{G} \in \hat{\omega}^{<\hat{\omega}}$ and write $\hat{n} = g/\dot{G} \in \hat{\omega}$. The choice of $g$ means that each $\hat{s}_\alpha \in \hat{n}^{<\hat{n}}$ (i.e., this is forced by $B$).
For each $\alpha < \theta$, consider $Y_\alpha$ as a function from $\omega$ to $[\omega]^{<\omega}$, given by $n \mapsto Y_{\alpha, n}$; let $\hat{a}_\alpha = Y_\alpha/\dot{G} \in [\hat{\omega}]^{<\hat{\omega}}$. Then $\hat{a}_\alpha = \{\hat{s} \in \hat{n}^{<\hat{n}}: \hat{s}_\alpha \subseteq \hat{s}\}$ and so now it is clear $\hat{a}_\alpha \supseteq \hat{a}_\beta$ for $\alpha < \beta$. Since this is forced by $B$, we get that $Y_\alpha \supseteq^* Y_\beta$ for $\alpha < \beta$.
Incidentally, the forcing extension is superficial. At the beginning of the proof we can suppose by passing to a Levy collapse that $\mathfrak{t} = 2^{\aleph_0}$. This allows us to build a sufficiently generic tower $(A_\gamma: \gamma < 2^{\aleph_0})$; the ultrafilter $\mathcal{U}$ generated by $(A_\gamma: \gamma < 2^{\aleph_0})$ will be as needed.
I'll close the question collecting here what helped me understanding the problem (thank you really much KP Hart and Douglas Ulrich for the help).
The thing that caused me great confusion is that I wasn't aware of the following fact: given a model $\mathfrak{M}$ and a $\lambda$-closed forcing notion $\mathbb{P}\in \mathfrak{M}$ and $G\subset \mathbb{P}$ ultrafilter generic over $\mathfrak{M}$, then $\mathfrak{M}[G]\vDash$ "$G$ (considered as a suborder of $\mathbb{P}$) is $\lambda$-closed".
I'll rewrite part of the statement of my example. I made a lot of confusion and used the wrong notation and symbols so I'll correct them here.
Let's consider $G$ an ultrafilter generic over $\mathfrak{M}$. $G$ need not to be fixed: it can be actually any of them.
Working in $\mathfrak{M}[G]$, find a subset $A\subset G$ such that $A$ is unbounded in $G$, that is $\forall x\in B[x\leq A\to x\notin G]$. Let $\dot{A}$ be the $B$-name of $A$ and $\dot{G}$ be the $B$-name of $G$ ($\check{x}$ should be used only for the canonical name of some $x\in\mathfrak{M}$, $G$ and $A$ does not belong to $\mathfrak{M}$). Then by forcing theorem, $\mathfrak{M}[G]\vDash$ "$A\subset G$ and $A$ is unbounded in $G$" if and only if there exists a $p\in G$ such that $p\Vdash$ "$A\subset G$ and $A$ is unbounded in $G$".
Here it might be useful to think $A=G$, and so we can take also $\dot{A}=\dot{G}$, since this satisfy all the requirement.
It is true that it is not possible that $\mathfrak{M}[G]\vDash p\leq A$, hence $\mathfrak{M}[G]\vDash p\nleq A$. For example in case $A=G$, $p$ might be the maximum of the boolean algebra, so this is not surprising.
That means $\mathfrak{M}[G]\vDash$ "there exists $a\in A$ such that $(p\land a)<p$", so we can fix an $a\in G$ and $p'\in G$ such that $(p\land a)<p$ and $p'\Vdash \check{a}\in \dot{A}$.
It is true that $((\neg a)\land p)>0$ and $((\neg a)\land p)<p$ and thus $((\neg a)\land p) \Vdash$ "$\dot{A}\subset \dot{G}$ and $\dot{A}$ is unbounded in $\dot{G}$", although a statement like "$((\neg a)\land p)\notin G$" makes sense only if we fixed $G$ so that $a\in G$, that is equivalent to say $a\Vdash ((\neg a)\land p)^{\check{}}\notin \dot{G}$.
So this alone does not create problem: it just means that $a$ and $p'$ incompatible and $p\not \Vdash \check{a}\in \dot{A}$, which is trivial. In fact, it's always true that given $q,r\in B$ if $q<r$ then $r\not \Vdash \check{q}\in \dot{G}$.
But in Fremlin's proof there were additional conditions that make this last stament less trivial (and hence surprising for me).
Assume $A$ wellordered and there exists $q\in B$, $\alpha$ ordinal and $b_i\in B$ for every $i<\alpha$ such that $q\Vdash$ "the order type of $\dot{A}$ is $\check{\alpha}$ and $\check{b_i}$ is the $i$-th element of $\dot{A}$ for any $i\in\check{\alpha}$".
Then we have a contradiction: in fact, since $q\in G$ and $q\Vdash$ "$\dot{A}$ unbounded in $\dot{G}$", we may find $a\in A$ such that $(q\land a)<q$ and $q'<q$ such that $q'\Vdash \check{a}\in\dot{A}$. Then there exists an $i\in\alpha$ such that $q'\Vdash \check{a}=\check{b_i}$ and so $a=b_i$, hence $q\Vdash \check{a}\in \dot{A}$. But now $((\neg a)\land q)>0$ and $((\neg a)\land q)<q$, hence $((\neg a)\land q)\Vdash \check{a}\in \dot{A}$ and at the same time $((\neg a)\land q)\Vdash \check{a}\notin \dot{A}$, contradiction.
This contradiction only proves that for every $A$ unbounded there is no such $q$. But if $\mathbb{P}$ is a $\lambda$-closed forcing notion such a $q$ can be found for every chain $A'$ of cardinality less than $\lambda$. This two things together were the cause of my confusion, while actually they just imply what stated at the beginning, i.e. that for every $\lambda$-closed forcing notion $\mathbb{P}$ and $G\subset \mathbb{P}$ generic ultrafilter over a model $\mathfrak{M}$, then $\mathfrak{M}[G]\vDash$ "$G$ is $\lambda$-closed".
My problem was that I didn't recognize what this contradiction meant at the beginning and thought there could have been generic ultrafilters with small unbounded chains, so I assumed $G$ fixed to avoid this contradiction, while actually it is not.