Let $H=(H, (\cdot, \cdot)_H)$ and $V=(V, (\cdot, \cdot)_V)$ be Hilbert spaces such that $V \hookrightarrow H$ that is $V$ is continuously embedding in $H$. By Riesz representation theorem we can identify $H=H'$. Thus we have $$V \hookrightarrow H=H' \hookrightarrow V'.$$
Since $H' \hookrightarrow V'$ then, in particular, $H' \subset V'$.
Question. If $f \in H'$ then we can to say that $f \in V'$?
I think so, in virtue of $H' \subset V'$. But more formally, due to $f \in H'$ then $f$ is linear. Now, I want to prove that $f$ is bounded (in $V'$). So I can to write, for every $v \in V$, $$\langle f , v \rangle _{V',V}= \langle f , v \rangle _{H',H} \leq \|f\|_{H'}\cdot \|v\|_{H} \leq \|f\|_{H'}\cdot c\cdot \|v\|_{V}?$$ Here, $c>0$ is the constant of the embedding $V \hookrightarrow H$.
The proof you have written works. Denoting the inclusion by $i:V\to H$ you have for $f\in H$ and $v\in V$ that:
$$|f(i(v))| ≤\|f\|_{H'} \cdot \|i(v)\|_H ≤ \|f\|_{H'}\cdot \|i\|_{L(V,H)}\cdot \|v\|_{V}$$
where you applied boundedness of $f$ and $i$ as linear maps to get that $v\mapsto f(i(v))$ is bounded as a linear map (with bound $\|f\|_{H'} \cdot \|i\|_{L(V,H)}$).
Depending on how precise you want to be it is not entirely correct to say that $f$ is an element of $V'$. Rather you should say that any $f\in H'$ induces an element of $V'$ by the prescription $f\mapsto f\circ i$.
Understanding the problem in that way also gives you a more immediate understanding why you get an element of $V'$, for $f\circ i:V\mapsto \Bbb K$ is continuous and linear as a composition of continuous linear maps.