Suppose that $S$, the relevant boundary component of $M$, is a torus. Suppose that $G$ is the given essential two-sphere in the filled manifold $N$. We isotope $G$ to have minimal intersection with $S$. All intersections $S \cap G$ are essential simple closed curves in $S$. So all of them are parallel. Set $F = M \cap G$. As you argue, $F$ is incompressible.

Suppose, for a contradiction, that $F$ is boundary compressible. Let $B \subset M$ be the given bigon boundary compressing $F$. Let $\beta = B \cap F$ and let $\beta' = B \cap S = B \cap \partial M$.

Let $\alpha$ and $\alpha'$ be the curves of $S \cap G$ which meet the corners of $B$ - that is, the points $\beta \cap \beta'$. If $\alpha = \alpha'$ then $\beta'$ is an inessential arc in $(S, \alpha)$. So $\beta'$ cuts a bigon out of $S - G$. Thus $B \cup B'$ is a compression of $F$, a contradiction.

Suppose instead that $\alpha$ and $\alpha'$ are distinct. Thus they co-bound an annulus $A$ in $S$. We choose $A$ so that $A \cap G = \alpha \cap \alpha'$. That is, there are no more curves of $G$ in $A$. Let $D$ be the disk obtained by boundary compressing $A$, along $B$, into $M$. This disk $D$ is a compressing disk for $F$ (or we can further reduce $G \cap S$). This is the desired contradiction.

If $S$ is not a torus, the argument is more involved. I will give details in my answer to your previous question.

Suppose that $S$, the relevant boundary component of $M$, is a torus. Suppose that $G$ is the given essential two-sphere in the filled manifold $N$. We isotope $G$ to have minimal intersection with $S$. All intersections $S \cap G$ are essential simple closed curves in $S$. So all of them are parallel. Set $F = M \cap G$. As you argue, $F$ is incompressible.

Suppose, for a contradiction, that $F$ is boundary compressible. Let $B \subset M$ be the given bigon boundary compressing $F$. Let $\beta = B \cap F$ and let $\beta' = B \cap S = B \cap \partial M$.

Let $\alpha$ and $\alpha'$ be the curves of $S \cap G$ which meet the corners of $B$ - that is, the points $\beta \cap \beta'$. If $\alpha = \alpha'$ then $\beta'$ is an inessential arc in $(S, \alpha)$. So $\beta'$ cuts a bigon out of $S - G$. Thus $B \cup B'$ is a compression of $F$, a contradiction.

Suppose instead that $\alpha$ and $\alpha'$ are distinct. Thus they co-bound an annulus $A$ in $S$. We choose $A$ so that $A \cap G = \alpha \cap \alpha'$. That is, there are no more curves of $G$ in $A$. Let $D$ be the disk obtained by boundary compressing $A$, along $B$, into $M$. This disk $D$ is a compressing disk for $F$ (or we can further reduce $G \cap S$). This is the desired contradiction.

If $S$ is not a torus, the argument is more involved. I have given the details for that in my answer to your previous question.