Interpreting $F[x,y]$ for $F$ a field. First, is it appropriate to write $F[x,y] = (F[x])[y] $? In particular, if $F$ is a field, then we know $F[x]$ is a Euclidean domain. Are there necessary and sufficient conditions for when $F[x,y]$ is also a Euclidean domain? 

In most constructions of polynomial rings (e.g., as almost null sequences with values in the ground ring), the rings $F[x,y]$, $(F[x])(y)$, and $(F[y])[x]$ are formally different objects: $F[x,y]$ is the set of all functions $m\colon \mathbb{N}\times\mathbb{N}\to F$ with $m(i,j)=0$ for almost all $(i,j)$; $(F[x])[y]$ is the set of all functons $f\colon\mathbb{N}\to F[x]$ with $f(n)=0$ for almost all $n$; and $(F[y])[x]$ is the set of all functions $g\colon\mathbb{N}\to F[y]$ with $g(m)=0$ for almost all $m$.
However, there are natural isomorphisms between them,
$$(F[x])[y]\cong F[x,y]\cong (F[y])[x].$$
Informally, this corresponds to the fact that we can write a polynomial in $x$ and $y$ by "putting $y$'s together" or by "putting $x$'s together. So, for instance, $1+x+2y + 3x^2y - 7xy^4 + 8x^2y^2$ can be written as
$$\begin{align*}
1+x+2y + 3x^2y - 7xy^4 + 8x^2y^2 &= 1 + (1-7y^4)x + (3y+8y^2)x^2\\
&= (1+x) + (2+3x^2)y + (8x^2)y^2 - (7x)y^4.
\end{align*}$$
So, yes, you can say that $F[x,y]$ is "essentially" equal to $(F[x])[y]$.
However, $F[x,y]$ is never a Euclidean domain, because it is never a PID: $\langle x,y\rangle$ is never principal.
In fact:
Theorem. Let $D$ be a domain. Then $D[x]$ is a PID if and only if $D$ is a field.
Proof. If $D$ is a field, then $D[x]$ is a Euclidean domain, hence a PID.
If $D$ is not a field, then let $a\in D$ be an element that is a nonzero nonunit. Then $\langle a,x\rangle$ is not a principal ideal: if $\langle a,x\rangle = \langle r\rangle$, then $r|a$, hence $r$ must be a polynomial of degree $0$; since $r|x$, then $r$ must be a unit or an associate of $x$; since it must be degree $0$, $r$ must be a unit. So $\langle a,x\rangle$ is principal if and only if it is equal to the entire ring. But $D[x]/\langle a,x\rangle \cong D/\langle a\rangle\neq 0$ (since $\langle a\rangle\neq D$), so $\langle a,x\rangle\neq D[x]$. Hence $\langle a,x\rangle$ is not principal. $\Box$
