Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$. If $f\in L^2(\Omega)$, then $$\langle u,\varphi\rangle:=\langle u,f\rangle_{L^2(\Omega)}\;\;\;\text{for }u\in H^1(\Omega)$$ is clearly a bounded linear functional on $H^1(\Omega)$. Now $H^1(\Omega)$ is a Hilbert space and hence, by Riesz' representation theorem, there is a $v\in H^1(\Omega)$ with $$\langle u,\varphi\rangle=\langle u,v\rangle_{H^1(\Omega)}=\langle u,v\rangle_{L^2(\Omega)}+\sum_{i=1}^d\langle\partial_iu,\partial_iv\rangle_{L^2(\Omega)}\tag1$$ for all $u\in H^1(\Omega)$.
Clearly, if $f\in H^1(\Omega)$, then $f=v$, but is there something we can generally say about the relation between $f$ and $v$, even when $f\not\in H^1(\Omega)$?
No, you do not have $f = v$, even in case $f \in H^1(\Omega)$. If this would be true, then $$ \langle u,f\rangle_{L^2} = \langle u,\varphi\rangle = \langle u, f\rangle_{L^2} + \sum_{i=1}^d \langle \partial_i u , \partial_i f\rangle_{L^2}$$ for all $u \in H^1$, which is clearly false (unless $\nabla f = 0$).
What is true, that $v \in H^1(\Omega)$ is the (weak) solution of $$ -\Delta v + v = f \text{ in } \Omega, \qquad\frac{\partial}{\partial n} v = 0 \text{ on } \partial\Omega.$$