The answer to question $1$ is no. Consider for example the ring $F[x,y]\big/(xy)$, for any field $F$. Then the natural map to $\left[F[x,y]\big/(x)\right]\times\left[F[x,y]\big/(y)\right]$ is not an isomorphism, as it does not take any element of $F[x,y]\big/(xy)$ to $(\overline{0},\overline{1})$ (can you see why?). However, the result does hold in the one-variable case, since $F[x]$ is a PID and hence a Bézout domain.
(Edit: To answer your question below, one can completely characterize the polynomials $f=f_1^{k_1}\dots f_n^{k_n}$ such that $F[x,y]\big/(f)\cong\prod_{i=1}^n F[x,y]\big/(f_i^{k_i})$, where the $f_i$ are non-associates and irreducible, as follows: the isomorphism holds if and only if $(f_i)+(f_j)=R$ for every $i\neq j$. For one direction, we may use the Chinese Remainder Theorem; since $F[x,y]$ is a UFD, $(f)=\bigcap_{i=1}^n(f_i^{k_i})$, and, if $(f_i)+(f_j)=R$ for every $i\neq j$, we have that the $(f_i^{k_i})$ are pairwise coprime, as needed. For the other direction, suppose $(f_i)+(f_j)\neq R$. Then no element of $F[x,y]\big/(f)$ maps to the element of $\prod_{i=1}^n F[x,y]\big/(f_i^{k_i})$ with a $\overline{1}$ in the $j$-th position and $\overline{0}$ everywhere else. Indeed, if $p$ were such an element, $p$ would be divisible by $f_i$ and congruent to $1$ mod $f_j^{k_j}$, and hence we would have $1\in (f_i)+(f_j)$, a contradiction. Thus $f$ induces the desired isomorphism if and only if $(f_i)+(f_j)=R$ for every $i\neq j$, and this gives a complete characterization.)
The answer to question $2$ depends very much on the polynomial $f$, and it's hard to say anything specific without more details. What kind of results are you looking for? For instance, we know that the ring $F[x,y]\big/(f)$ will be an integral domain of Krull dimension $1$, but I'm not sure if this is the kind of thing you're looking for.
The answer to question $3$ is yes. Indeed, we can say something more general; let $F$ be any field, and let $p\in F[x]$ be any polynomial, and consider the map $f:F[x,y]\to F[x]$ defined by $x\mapsto x$ and $y\mapsto p$. Clearly $f$ is surjective, and we claim $\ker f=(p-y)$. The $\supseteq$ inclusion is clear, and for the $\subseteq$ inclusion suppose $f(q)=0$ for some $q\in F[x,y]$. Then in particular $q(x,p)=0\in F[x]\subseteq F[x,y]$. On the other hand, replacing every instance of $\overline{y}$ in $\overline{q(x,y)}$ by $\overline{p}=\overline{y}$ gives $$\overline{q(x,y)}=\overline{q(x,p)}=\overline{0}\in F[x,y]\big/(p-y),$$ and so $q(x,y)\in (p-y)$, as desired. Thus, by the first isomorphism theorem, $f$ induces the desired isomorphism $F[x,y]\big/(p-y)\cong F[x]$.
