I think what you have is correct, but it's a bit hard to follow.

First, you don't really define $f^{\ast}$. I think what you probably intend is:

$f^{\ast}:W^{\ast} \to V^{\ast}$ given by, for $\varphi \in W^{\ast}$: $f^{\ast}(\varphi) = \varphi \circ f$.

Let's see if we can provide a more direct proof:

Suppose $f$ is injective, and let $\psi$ be any element of $V^{\ast}$.

Since $f$ is injective, there is $g \in \text{Hom}_{\ F}(\text{im }f,V)$ with $g \circ f = 1_V$.

Now consider $g^{\ast}:V^{\ast} \to (\text{im }f)^{\ast}$.

By definition: $(f^{\ast}\circ g^{\ast})(\psi) = f^{\ast}(g^{\ast}(\psi)) = (g^{\ast}(\psi)) \circ f = (\psi \circ g) \circ f = \psi \circ (g \circ f) = \psi \circ 1_V = \psi$.

Since $g^{\ast}(\psi) \in (\text{im }f)^{\ast} \subseteq W^{\ast}$, it follows that $g^{\ast}(\psi)$ is an element $\varphi \in W^{\ast}$ with $f^{\ast}(\varphi) = \psi$, that is, $f^{\ast}$ is surjective.

On the other hand, suppose $f^{\ast}$ is surjective.

This, in turn, means that there is $g^{\ast}: V^{\ast} \to W^{\ast}$ such that $f^{\ast} \circ g^{\ast} = 1_{V^{\ast}}$.

So for any $\psi \in V^{\ast}$, we have:

$\psi = 1_{V^{\ast}}(\psi) = (f^{\ast} \circ g^{\ast})(\psi) = f^{\ast}(g^{\ast}(\psi)) = (g^{\ast}(\psi)) \circ f = (\psi \circ g) \circ f = \psi \circ (g \circ f)$.

In particular, for any $v \in V$, with $\psi = \langle v,-\rangle$, we have $\langle v,-\rangle = \langle (g \circ f)(v),-\rangle$ so that:

$\langle v - (g \circ f)(v),x\rangle = 0$ for all $x \in V$, including $x = v - (g \circ f)(v)$, so that $v = (g \circ f)(v)$ for all $v \in V$, that is $g \circ f = 1_V$, and $f$ is injective.