Let $K$ be a field, and $X/K$ be an affine scheme, say given by equations $f_1＝f_2＝\cdots＝f_r＝0$.
Why is the set $$\{K\text{-morpisms } \operatorname{Spec}K \to X\}$$ in bijection with $$\{K\text{-algebra homomorphisms } K[x_1,x_2,\cdots,x_n]/(f_1,f_2,\cdots,f_r)\to K\}$$ ?
I think this is one of the most basic fact in algebraic geometry, but I'm not familiar with algebraic geometry, so a self-contained explanation is appreciated. Thank you in advance.
$\operatorname{Spec}$ is a contravariant equivalence of categories between the categories of commutative rings and affine schemes. In your situation, this means that $K$-morphisms $$\operatorname{Spec} K\to X=\operatorname{Spec} K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$$ are in bijection with $K$-algebra homomorphisms $K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)\to K$.
You can find the first statement in any (introductory) algebraic geometry book. This gives that morphisms $\operatorname{Spec} K\to X$ are in bijection with maps of rings $K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)\to K$ (because $X=\operatorname{Spec} K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$). The condition that these are $K$-maps then enforces that these maps of rings should be $K$-algebra homomorphisms: a $K$-scheme is a scheme that comes with a morphism to $\operatorname{Spec} K$ often called the structure map, and we say a map of $K$-schemes is a $K$-map if it fits in to a commutative triangle
$$\begin{array}{ccc} X & \xrightarrow{} & Y \\ & \searrow & \downarrow \\ & & \operatorname{Spec} K \end{array}$$
where the down and diagonal arrows are the structure maps. For $\operatorname{Spec} K$, the structure map is the identity, and for $X=\operatorname{Spec} K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$ the structure map is $\operatorname{Spec}$ of the $K$-linear map taking $1\mapsto 1$. So a $K$-map $\operatorname{Spec} K\to X$ is equivalent to a ring homomorphism $K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)\to K$ which when pre-composed with the usual inclusion $K\to K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$ gives the identity on $K$. This is exactly a $K$-algebra homomorphism.
Finally, a note of caution: attempting to dig in to Neron models (as in your other algebraic-geometry questions) before knowing something as basic as the fact that $\operatorname{Spec}$ is an equivalence is a bit like trying to qualify for a race in the Olympics before you know how to walk. I would suggest you seriously consider taking some time to understand a bit more of the basics of the field.