A morphism $F$ between algebraic varieties $V\subseteq \Bbb{A}^n$ and $W\subseteq \Bbb{A}^m$ is the restriction of a polynomial map on the ambient affine spaces $\Bbb{A}^m$ and $\Bbb{A}^n$.
What does this mean? Say we have the mapping $F:\Bbb{A}^2\to\Bbb{A}^3$ defined by $(x,y)\to (x+y,x^2+y^2,x^3+y^3)$. Also, we choose $\Bbb{V}(x-y)$ to be the variety in $\Bbb{A}^2$ and $\Bbb{V}(x+2y+3z)$ in $\Bbb{A}^3$. Does $F$ necessarily map $\Bbb{V}(x-y)$ to $\Bbb{V}(x+2y+3z)$? What is happening?
Let me just set up notation again. I've got $F\colon \mathbb{A}^n \to \mathbb{A}^m$ and affine varieties $V \subset \mathbb{A}^n$ and $W \subset \mathbb{A}^m$. When is it the case that $p \in V$ implies $F(p) \in W$? The condition to be in $W$ is that $g(F(p)) = 0$ for each $g$ in some set of polynomials cutting out $W$, so I want the polynomial function $g \circ F$ to vanish on $V$; ie, it should lie in $I(V)$.
In your example it might be better to give different names to the coordinates on the source and target, but forging on we check that $x + 2y + 3z$ pulls back to \[ (x + y) + 2(x^2 + y^2) + 3(x^3 + y^3). \] This is not a multiple of the generator $x - y$ of $I(V)$. Of course, just checking that $F(1, 1) = (2, 2, 2)$ doesn't satisfy $x + 2y + 3z$ is a fine argument as well.