Mariano Suarez-Alvarez's point about understanding the intuition as you learn the theory more is correct, but I'd like to give a partial answer to help guide your intuition. After all, it is possible to spend months or years learning algebraic geometry and come away with little intuition of what the whole subject is about.
First, algebraic varieties are geometric spaces which look locally like affine varieties. In this sense, the theory is developed similar to, say, the theory of manifolds where a manifold is defined to be a space that is locally Euclidean. Of course, that limits the local study of manifolds - any two manifolds are locally isomorphic. Not so for algebraic varieties, as there is a wide variety of affine varieties.
So I think you should begun by restricting your question to affine varieties. And the key is that affine varieties are completely determined by their ring of globally regular functions. In other words, two (irreducible) closed subsets of affine space are isomorphic iff we can find a global 'change of variables' that identifies the global regular functions on the two spaces. Rescaling $(x,y) \mapsto (\sqrt{2}x,\sqrt{2}y)$ yields the isomorphism between $x^2+y^2=1$ and $x^2+y^2=2$.
I'll modify your non-example (because $\mathbb{A}^2 \setminus \{0\}$ is not affine) and explain why $\mathbb{A}^1$ and $\mathbb{A}^1 \setminus \{0\}$ are not isomorphic. Their rings of regular functions are $k[T]$ and $k[T,T^{-1}]$ respectively, which are not isomorphic. So there can be no 'changes of variables' that identifies the two spaces.
One important caveat: when I say there is global 'change of variables' from $X \subset \mathbb{A}^n$ and $X' \subset \mathbb{A}^{n'}$, I am talking about using polynomial maps that are restricted from the respective affine spaces, but they only needed to be defined on the spaces $X$ and $X'$. For example $\mathbb{A}^1 \setminus \{0\}$ (viewed as $t \neq 0$) and $xy=1$ are isomorphic via $t \mapsto (t, 1/t)$ and $(x,y) \mapsto x$. Of course, $1/t$ is only a valid change of variables when $t \neq 0$, but fortunately we are only looking at points where $t \neq 0$.
The global story is similar, except that we cannot just compare globally regular functions. (For example, the only globally regular functions on any projective variety are the constant functions, yet intuitvely there ought to be many different projective varieties up to isomorphism.) So now we require a global 'change of variables' so that regular functions on local pieces match up with the regular functions on the corresponding local pieces.
I am not sure if this explanation is what you are looking for. Algebraic geometry is very much a function oriented theory. We compare spaces by looking at the functions on them. One can take such an approach to manifolds as well. But for manifolds we also have an intuition for what the possible change of variables are ('stretching' and 'twisting' and the like). It's much harder to tell such a story in algebraic geometry because algebraic varieties are so much more diverse. There are still some basic intutions such as you can't have an isomorphism between a smooth variety and a singular variety because isomorphisms give rise to (vector space) isomorphisms of tangent spaces. But there are lots of possible singularities, and getting a hold on them is a major on-going project in the field. For example, you could study plane curves in depth and learn to tell apart singularities in this case (using blowups). But then you'll quickly discover the singularities on surfaces are more complicated and those on higher dimensional varieties still more complicated and hard to get a handle on.
It's true that an algebraic map of Zariski-closed subsets of $K^n$ (this is not what "affine variety over $K$" means unless $K$ is algebraically closed) is an isomorphism iff the induced map on rings of functions is an isomorphism. The point is that an inverse map on rings of functions provides the components of an algebraic map which inverts the original map.
Depending on what you mean by $\text{Spec }$, though, the last part of the equivalence does not hold. Let me assume for simplicity that $K$ is algebraically closed. Consider the map
$$\mathbb{A}^1 \ni t \mapsto (t^2, t^3) \in \mathbb{A}^2.$$
The image of this map is the variety $\{ y^2 = x^3 \}$, with ring of functions $K[x, y]/(y^2 - x^3)$. The induced map
$$K[x, y]/(y^2 - x^3) \ni f(x, y) \mapsto f(t^2, t^3) \in K[t]$$
on rings of functions is not an isomorphism because its image does not contain $t$. But the induced induced map on prime ideals is an isomorphism (of sets).
From a differential-geometric point of view, the problem is that the map $t \mapsto (t^2, t^3)$ does not induce an isomorphism on Zariski tangent spaces at the origin; that is, it is neither an immersion nor a submersion at the origin. This discrepancy cannot be detected at the level of prime ideals, but it can be detected using a non-prime ideal, namely the ideal $(t^2)$.
If you haven't already, it would be good at this point to spend some time studying the Nullstellensatz.
Best Answer
The final part of the question, as amplified in Mohan's comment, has an affirmative answer. $\mathbb A^1-\{0\}$ is open and not closed in $\mathbb A^1$ but its image, under the isomorphic embedding $x\mapsto(x,0)$ is closed and not open in $\mathbb A^2-\{(0,0)\}$.
As for the earlier parts of the question, isomorphism means that the two algebraic sets are not merely in one-to-one correspondence any old way but rather in a way that respects the algebraic structure. By "the algebraic structure" here, I mean the information telling which functions on the algebraic set are polynomials. The definition of isomorphism ensures that, if you have a polynomial function on one of the two isomorphic algebraic sets, then transporting that function to the other set, by composing with the bijection $\phi$ or its inverse, produces a polynomial function there. (Both my explanation and the definition in the question seem to need some caution in finite characteristic, where a polynomial isn't determined by its values.)