Where $V\left(f\right)=\left\{a\in \mathbb{K}^n\::\:f\left(a\right)=0\right\}$ and $f\in \:\mathbb{K}\left[X_1,\:...,\:X_n\right]$
I don't understand why
$$\left\{p\:\in \:Spec\:R\::\:f\:\notin \:p\right\}=Spec\:R\:\setminus \:V\left(f\right)$$
I just can't see it. Could anyone maybe explain this?
The main problem here is that you've confused the "classical" and "modern" Zariski topologies! Let me start by explaining the difference between these two things. Note: this is not really standard terminology, but I don't have a better way to disambiguate these two versions of the theory, so I'll keep saying "classical" and "modern" throughout this answer.
The classical Zariski topology is a topology on $\mathbb{K}^n$, where $\mathbb{K}$ is an algebraically closed field. The basic closed sets are precisely the sets $V(f) = \{a \in \mathbb{K}^n : f(a) = 0\}$ you defined. Thus, in the classical setting, the basic opens are of the form $D(f) := \mathbb{K}^n \setminus V(f) = \{a \in \mathbb{K}^n : f(a) \neq 0\}$. There is absolutely nothing to prove here, these are just definitions.
The modern Zariski topology is a topology on $\operatorname{Spec} R$, where $R$ is a commutative ring. This is never the same set as $\mathbb{K}^n$, so the $V(f)$'s that you defined are not even subsets of $\operatorname{Spec} R$. Instead, the basic closed sets in the modern Zariski topology are $$V(f) = \{\mathfrak{p} \in \operatorname{Spec}(R) : f \in \mathfrak{p}\}$$ where $f \in R$. Again: these are not the same as the $V(f)$'s in the classical setting! The modern basic opens $D(f) = \{\mathfrak{p} \in \operatorname{Spec}(R) : f \notin \mathfrak{p}\}$ are precisely the complements of the modern basic closeds. There is absolutely nothing to prove here, these are just definitions.
So what's the connection between the classical and modern Zariski topologies? Well, let $R = \mathbb{K}[X_1, \dots, X_n]$ where $\mathbb{K}$ is an algebraically closed field. By the nullstellensatz, the maximal ideals of $R$ are precisely the ideals $\mathfrak{m}_{(a_1, \dots, a_n)} = (X_1 - a_1, X_2 - a_2, \dots, X_n - a_n)$ as $(a_1, \dots, a_n)$ ranges over all points in $\mathbb{K}^n$. In $\operatorname{Spec}(R)$, the maximal ideals are precisely the closed points, so we have a bijective correspondence $$\{\text{closed points of }\operatorname{Spec} \mathbb{K}[X_1, \dots, X_n]\} \leftrightarrow \mathbb{K}^n$$ What's more is that the subspace topology on $\{\text{closed points of }\operatorname{Spec} \mathbb{K}[X_1, \dots, X_n]\}$ makes this bijection a homeomorphism! This is a very important exercise (and can be done using only the information I expounded in this answer). In particular, the classical basic closed set $V(f)$ corresponds to $\{\mathfrak{m}_a : f(a) = 0\}$, which is the intersection of the set of closed points with the modern version of $V(f)$ -- you should also prove this.