Well the standard answer to this sort of question is that two algebraic objects (vector spaces in this case) $V$ and $W$ are isomorphic if they are basically the same, meaning that once can identify them with one another in a reasonable way. Another way to say this is that a map $f:V\to W$ is an isomorphism if it is bijective and it preserves the algebraic structures of $V$ (or $W$, since the definition implies that $f^{-1}:W\to V$ is an isomorphism).
In the case of vector spaces, the algebraic structure we're interested in is addition of vectors and multiplication by scalars, so that's the basis (no pun intended) for the definition $f(av_1+bv_2)=af(v_1)+bf(v_2)$ in the definition of a homomorphism (linear transformation) of vector spaces).
Why did I say basically the same and not exactly the same? The typical example here would be $V=\mathbb{R}^2$ and $W=\mathbb{C}$ (over $\mathbb{R}$). These two vector spaces aren't exactly the same, since $\mathbb{C}$ has a lot of different algebraic properties than $\mathbb{R}^2$. However, as vector spaces, the map taking $(1,0)\mapsto 1$ and $(0,1)\mapsto i$ is an isomorphism (as you can check), so as real vector spaces, they are essentially the same.
Another note here is that we used a specific choice of basis in this example for the isomorphism. There are spaces which are canonically isomorphic, which essentially means we can create an isomorphism between them that doesn't depend on the choice of basis. I can't think of an elementary linear algebra example right now other than $V$ and $V^{\ast\ast}$, the double dual space, are naturally isomorphic when $V$ is finite dimensional. Still, while basically the same, they are not exactly the same: one is a vector space of just abstract vectors, while the other is a space of functions from $V^{\ast}\to \mathbb{F}$ (the base field).
Hope that helps.