Show that If a sequence is exact the dual sequence is exact

$\newcommand{\im}{\operatorname{im}}$Suppose that $\varphi \in V^*$ is in the kernel of $f^*$. Since $f^*\varphi = \varphi \circ f = 0$, we know that $\im f \subseteq \ker \varphi$. Since the original sequence is exact, $\im f = \ker g$. So by the Fundamental Theorem on Homomorphisms (applied to $\varphi$), there is a unique map $\tilde\varphi \colon V/\ker g\to K$ such that $\tilde\varphi\circ \pi = \varphi$, where $\pi$ is the canonical epimorphism from $V$ to $V/\ker g$.

By the same theorem applied to $g_1$, there is a unique map (in fact, an isomorphism) $\tilde g_1 \colon V/\ker g \to \im g$ such that $\tilde g_1 \circ \pi = g_1$. Let $\varphi' = \tilde\varphi \circ \tilde g_1 ^{-1} \colon \im g \to K$. Then $$ \varphi' \circ g_1 = \tilde\varphi\circ\tilde g_1^{-1} \circ g_1 = \tilde\varphi\circ\pi = \varphi $$ as desired.

For the rest of the construction, you need to extend $\varphi'$ to be defined on all of $W$ instead of just $\im g$. There isn't a canonical way to do this without extra structure on $W$, but there are many choices.