Duality: dual maps and linear maps

Suppose we have $$V$$ and $$W$$ be $$2$$ $$K$$-vectorspaces and $$f:V\rightarrow W$$ is a linear map then:

1. For all $$\varphi\in W^\ast$$ one has $$\varphi\circ f\in V^*$$
2. The map $$f^\ast:W^\ast\rightarrow V^\ast:\varphi\mapsto\varphi\circ f\;$$ is a linear map, which is also called a dual map.

Where $$V^\ast$$, the dual space, is defined as: $$$$V^\ast:= Hom_K(V,K) = \{\;f:V\rightarrow K\;|\;f\;\text{is a linear map}\;\}$$$$

Where $$K$$ is a $$1$$-dimensional $$K$$-vector space and $$V$$ a $$K$$-vector space

But I don't really get what this means, could anyone give a clear explanation because I have a hard time understanding this. Especially with that extra $$\varphi$$ function

The set $$V^*$$ of all linear maps from a vector space $$V$$ to its base field $$K$$ is, itself, a vector space over $$K$$: this is easy to show, with the operations being $$(f+g)(x) = f(x) + g(x)$$ and $$(\lambda f)(x) = \lambda(f(x))$$, for all $$f, g\in V^*$$ and all $$\lambda \in K$$.

Now, given any $$\varphi \in W^*$$, and any $$f: V \to W$$, $$\varphi\circ f$$ is a linear map from $$V$$ to $$K$$ (since it's a composition of linear maps), so is an element of $$V^*$$ (you have a typo in your question: it is not an element of $$V$$).

Thus, we define a new function, $$f^*$$, that takes as its input an element $$\varphi$$ of $$W^*$$, and spits out an element of $$V^*$$, by composing $$\varphi$$ with $$f$$. This, it turns out, is linear, and this, too, is not hard to check: if $$\varphi,\psi\in W^*$$, and $$v \in V$$, then

\begin{align*}f^*(\varphi+\psi)(v) &= (\varphi + \psi)\circ f(v) \\&= (\varphi+\psi)(f(v)) \\&= \varphi(f(v)) + \psi(f(v)) \\&= \varphi\circ f(v) + \psi \circ f(v) \\&= (f^*\varphi + f^*\psi)(v),\end{align*}

hence $$f^*(\varphi + \psi) = f^*\varphi + f^*\psi$$. Similarly, for any $$\lambda \in K$$, we have

\begin{align*}f^*(\lambda\varphi)(v) &= (\lambda\varphi)\circ f(v) \\&= (\lambda\varphi)(f(v))\\&=\lambda(\varphi(f(v))) \\&= \lambda(\varphi \circ f)(v)) = \lambda f^*(\varphi)(v),\end{align*}

hence $$f^*(\lambda\varphi) = \lambda f^*(\varphi)$$, and $$f^*$$ is, indeed, linear.