I've read from some lecture notes that the idea to define morphisms is that a variety $X$ is completely determined by the regular functions on all open subsets of $X$, so a morphism should preserve regular functions on open subsets.
How can I see that a variety is completely determined by the regular functions on all open subsets?
By definition a variety is an irreducible algebraic set, so it is the zero set of some polynomials. What is the relation between this and the regular functions?
Let $V \subseteq \mathbb{A}^n$ be a variety. Let $I$ be the ideal corresponding to $V$, and let $R$ be the polynomial ring $k[X_1, ... , X_n]$. Then $V$ is completely determined by the ring $k[V] = R/I$: the points of $V$ can be identified with the maximal ideals of $R/I$, and one can define the Zariski toplogy, regular functions etc. directly on this space of maximal ideals.
For $U \subseteq V$ open, we have the ring of regular functions $\mathcal O_V(U)$, consisting of all functions $U \rightarrow k$ which locally can be represented as a quotient of polynomial functions.
Actually, a variety is completely determined just by the ring of global functions $\mathcal O_V(V)$, because the natural homomorphism $k[V] \rightarrow \mathcal O_V(V)$ is an isomorphism. See the first chapter of Linear Algebraic Groups by T.A. Springer for a straightforward approach, or section 1.3 of Hartshorne for a more slick proof.