We often defines elliptic curve over $K$ as an （projective closure of）affine curve defined by Weierstrass equation, $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6\text{ over a field K.}$$
In this context, x,y is element of $K$, because this is curve in affine space over $K$.
But we sometimes say $x$ has order $2$,and $y$ has order $2$. In this context, we should see $x$ and $y$ as elements of $K(E)$.
I think this is misleading,
・Why can we define order of $x$,$y$ of weierstrass equation?
・When we write Weierstrass equation, $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6\text{ over a field K.}$$ Which set we should think $x$,$y$ belongs to ?
$f(x,y)=0$ defines the coordinate ring $K[E]=K[x,y]/(f(x,y))$ and every $a,b\in \overline{K}$ such that $f(a,b)=0$ corresponds to an homomorphism $K[E]\to \overline{K}$.
$x,y$ are elements of the coordinate ring, and its field of fractions $K(E)=K(x)[y]/(f(x,y))$.
On $K[E]$ and $K(E)$ we can define the order of zero and pole at such a $(a,b)$, because it is a smooth curve. Its projective closure is also smooth at the point at infinity so we can define the order of zero and pole at every point. In general for any curve $C$ we can also define some order of zero/poles directly from discrete valuations on the function field $K(C)$, which is quite the same as finding a smooth projective curve whose function field is $K(C)$.
With the order stuff there is another story: if a smooth projective curve $C$ is such that there is $P\in C$ and two elements $u,v$ of its function field $K(C)$ whose only poles are double and triple at $P$, and no element of $K(C)$ has only one simple pole at $P$, then $u,v$ satisfy a Weierstrass equation $f(u,v)=0$ and $C$ is (isomorphic to...) the elliptic curve $f(u,v)=0$.