1. First note that $\lnot A \vee B$ is tautologically equivalent to $A \rightarrow B$ so: $1 \Vdash A \rightarrow B$ $\Leftrightarrow$ $1 \Vdash \lnot A \vee B$. Specifically, $A \rightarrow B$ will be true in every forcing extension if and only if $\lnot A \vee B$ is true in every forcing extension.

2. But to show $1 \Vdash \lnot A \vee B$, it is sufficient to show that the set $D$ of conditions from $P$ forcing $\lnot A \vee B$ is dense in $P$. This is because if $G$ is a $V$-generic filter for $P$, then it must meet every dense set. Consequently, if $D$ is dense, then there will be a condition $p$ in $G \cap D$. In this case, $p \in G$ and $p \Vdash \lnot A \lor B$ so $V[G] \models \lnot A \lor B$, or equivalently, $V[G] \models A \rightarrow B$.

3. Fix a $q$ in $P$. If any $p \lt q$ forces $\lnot A$, then it will also force $\lnot A \lor B$ because $\lnot A \vee B$ is a tautological consequence of $\lnot A$. Specifically, if every forcing extension $V[G]$ such that $p \in G$ models $\lnot A$, then every such extension will also model $\lnot A \vee B$.

4. We may therefore assume that all $p \lt q$ do not force $\lnot A$ so that $q \Vdash A$. Notice that if $q$ did not force $A$, then we'd have $V[G] \models \lnot A$ for some $G$ where $q \in G$. In this case, we would have some condition $r \in G$ forcing $\lnot A$. Now notice that if $r \Vdash \lnot A$ and $s < r$, then also $s \Vdash \lnot A$ since every filter containing $s$ will contain $r$ by the upward closure of filters. But also if $q \in G$ and $r \in G$, then we may find such a condition $s$ in $G$ below both $q$ and $r$ by virtue of $G$ being a filter. This $s$ will then be below $q$ and will force $\lnot A$ by virtue of being less than $r$, contrary to assumption.

5. It therefore suffices to show that we have a condition $p$ below $q$ forcing $B$. If we let $q \in G$, and $V[G] \models B$, then we will have a condition $r \Vdash B$ such that $r \in G$. As in 4., we may then find an $s \in G$ below $q$ that forces $B$ (and hence also $\lnot A \lor B$), and so we are done.

1. First note that $\lnot A \vee B$ is tautologically equivalent to $A \rightarrow B$ so: $1 \Vdash A \rightarrow B$ $\Leftrightarrow$ $1 \Vdash \lnot A \vee B$. Specifically, $A \rightarrow B$ will be true in every forcing extension if and only if $\lnot A \vee B$ is true in every forcing extension.

2. But to show $1 \Vdash \lnot A \vee B$, it is sufficient to show that the set $D$ of conditions from $P$ forcing $\lnot A \vee B$ is dense in $P$. This is because if $G$ is a $V$-generic filter for $P$, then it must meet every dense set. Consequently, if $D$ is dense, then there will be a condition $p$ in $G \cap D$. In this case, $p \in G$ and $p \Vdash \lnot A \lor B$ so $V[G] \models \lnot A \lor B$, or equivalently, $V[G] \models A \rightarrow B$.

3. Fix a $q$ in $P$. If any $p \lt q$ forces $\lnot A$, then it will also force $\lnot A \lor B$ because $\lnot A \vee B$ is a tautological consequence of $\lnot A$. Specifically, if every forcing extension $V[G]$ such that $p \in G$ models $\lnot A$, then every such extension will also model $\lnot A \vee B$.

4. We may therefore assume that all $p \lt q$ do not force $\lnot A$ so that $q \Vdash A$. Notice that if $q$ did not force $A$, then we'd have $V[G] \models \lnot A$ for some $G$ where $q \in G$. In this case, we would have some condition $r \in G$ forcing $\lnot A$. Now notice that if $r \Vdash \lnot A$ and $s < r$, then also $s \Vdash \lnot A$ since every filter containing $s$ will contain $r$ by the upward closure of filters. But also if $q \in G$ and $r \in G$, then we may find such a condition $s$ in $G$ below both $q$ and $r$ by virtue of $G$ being a filter. This $s$ will then be below $q$ and will force $\lnot A$ by virtue of being less than $r$, contrary to assumption.

5. It therefore suffices to show that we have a condition $p$ below $q$ forcing $B$. If we let $q \in G$, and $V[G] \models B$, then we will have a condition $r \Vdash B$ such that $r \in G$. As in 4., we may then find an $s \in G$ below $q$ that forces $B$ (and hence also $\lnot A \lor B$), and so we are done.

1. No, if $A$ holds in a forcing extension $V[G]$, it need not be forced by $1$ in general. But this is not what is done. Instead, the argument can proceed as follows: In order to show that $1$ forces a statement to be true, we may show it is true in all forcing extensions. Consider an arbitrary forcing extension $V[G]$. If $V[G] \nvDash A$, then $V[G] \models A \rightarrow B$. Consequently, it suffices to restrict our attention to forcing extensions $V[G]$ for which $A$ is true and then show that $B$ will also be true in $V[G]$.

2. I think this question got cut off somehow, and I'm not sure what the question was going to be. I'll answer after you edit it, or most likely someone else will respond before I do.

1. No, if $A$ holds in a forcing extension $V[G]$, it need not be forced by $1$ in general. But this is not what is done. Instead, the argument can proceed as follows: In order to show that $1$ forces a statement to be true, we may show it is true in all forcing extensions. Consider an arbitrary forcing extension $V[G]$. If $V[G] \nvDash A$, then $V[G] \models A \rightarrow B$. Consequently, it suffices to restrict our attention to forcing extensions $V[G]$ for which $A$ is true and then show that $B$ will also be true in $V[G]$.

2. I'm still not sure which part of the proof you find problematic, but let me try to break it down step by step.

1. First note that $\lnot A \vee B$ is tautologically equivalent to $A \rightarrow B$ so: $1 \Vdash A \rightarrow B$ $\Leftrightarrow$ $1 \Vdash \lnot A \vee B$. Specifically, $A \rightarrow B$ will be true in every forcing extension if and only if $\lnot A \vee B$ is true in every forcing extension.

2. But to show $1 \Vdash \lnot A \vee B$, it is sufficient to show that the set $D$ of conditions from $P$ forcing $\lnot A \vee B$ is dense in $P$. This is because if $G$ is a $V$-generic filter for $P$, then it must meet every dense set. Consequently, if $D$ is dense, then there will be a condition $p$ in $G \cap D$. In this case, $p \in G$ and $p \Vdash \lnot A \lor B$ so $V[G] \models \lnot A \lor B$, or equivalently, $V[G] \models A \rightarrow B$.

3. Fix a $q$ in $P$. If any $p \lt q$ forces $\lnot A$, then it will also force $\lnot A \lor B$ because $\lnot A \vee B$ is a tautological consequence of $\lnot A$. Specifically, if every forcing extension $V[G]$ such that $p \in G$ models $\lnot A$, then every such extension will also model $\lnot A \vee B$.

4. We may therefore assume that all $p \lt q$ do not force $\lnot A$ so that $q \Vdash A$. Notice that if $q$ did not force $A$, then we'd have $V[G] \models \lnot A$ for some $G$ where $q \in G$. In this case, we would have some condition $r \in G$ forcing $\lnot A$. Now notice that if $r \Vdash \lnot A$ and $s < r$, then also $s \Vdash \lnot A$ since every filter containing $s$ will contain $r$ by the upward closure of filters. But also if $q \in G$ and $r \in G$, then we may find such a condition $s$ in $G$ below both $q$ and $r$ by virtue of $G$ being a filter. This $s$ will then be below $q$ and will force $\lnot A$ by virtue of being less than $r$, contrary to assumption.

5. It therefore suffices to show that we have a condition $p$ below $q$ forcing $B$. If we let $q \in G$, and $V[G] \models B$, then we will have a condition $r \Vdash B$ such that $r \in G$. As in 4., we may then find an $s \in G$ below $q$ that forces $B$ (and hence also $\lnot A \lor B$), and so we are done.

As a technical point, it is implicitly assumed throughout this argument that there is no minimal condition in $P$. Otherwise, replace $\lt$ with $\leq$.