Clearly $f\in V_\phi$ iff $\hat{f}(\gamma) = p(\gamma)\hat{\phi}(\gamma)$ for some $1$-periodic $p\in L^2(\mathbb{T})$, where $\hat{\phi}$ denotes the Fourier transform of $\phi$.
Let $$\Phi(\gamma) = \sum_{k\in\mathbb{Z}}|\hat{\phi}(\gamma-k)|^2 .$$
$\Phi\in L^1(\mathbb{T})$ since $\phi\in L^2(\mathbb{R})$.
Lemma: a. If $V_{\phi}$ is well-defined, then $\Phi\in L^{\infty}(\mathbb{T})$.
$\hspace{3mm}$ b. $V_\phi$ is isometrically isomorphic to $\{p\in L^2(\mathbb{T}): |p|^2\Phi\in L^1(\mathbb{T})\}$, a subspace of the weighted $L^2$-space $L^2_\Phi(\mathbb{T})$.
Proof. By Plancherel theorem, for all $f\in V_\phi$
\begin{eqnarray}
\|f\|_2^2 = \|\hat{f}\|_2^2
= \sum_{k\in\mathbb{Z}}\int_k^{k+1} |p(\gamma)\hat{\phi}(\gamma)|^2\ d\gamma
= \int_0^1 |p(\gamma)|^2\Phi(\gamma)\ d\gamma
\end{eqnarray}
If $\Phi\notin L^{\infty}(\mathbb{T})$ then there exists $p\in L^2(\mathbb{T})$ such that the integral on the RHS above diverges. Let $(c_n)$ be the Fourier coefficients of $p$. The series $\sum c_n\phi(\cdot - n)$ doesn't converge.
Proposition: Let $U:\overline{V_\phi}\to L^2_\Phi(\mathbb{T})$ be the isometry in the previous Lemma. The following are equivalent.
i. $V_{\phi}$ is closed in $L^2(\mathbb{R})$.
ii. $U(V_\phi) = \{p\in L^2(\mathbb{T}): |p|^2\Phi\in L^1(\mathbb{T})\}$ is closed in $L^2_\Phi(\mathbb{T})$.
iii. There exists $Z\subseteq\mathbb{T}$ and $A,B>0$ such that $\Phi(Z)=\{0\}$ and
$$A\leq \Phi(\gamma)\leq B \hspace{8mm} \forall\gamma\in\mathbb{T}\backslash Z \hspace{3mm}\textrm{a.e.}$$
Proof. $(i\Leftrightarrow ii)$ clear.
$\hspace{4mm}(iii\Rightarrow ii)$ $U(V_\phi)$ is isomorphic to $L^2(\mathbb{T}\backslash Z)$ so closed.
$(ii\Rightarrow iii)$ Let $\displaystyle A_n = \{\gamma\in \mathbb{T}\backslash Z : \frac{1}{2^{n+1}}\leq\Phi(\gamma)\leq\frac{1}{2^n} \}$.
Suppose $(iii)$ is false. Then, there exists a subsequence $(A_{n_k})$ with $m(A_{n_k})>0$. Let $$q_n = \sum_{k=1}^n \sqrt{\frac{k}{m(A_{n_k})}} \mathbf{1}_{A_k}$$ then $(q_n)$ is Cauchy in $U(V_{\phi})$, and converges in $L^2_\Phi(\mathbb{T})$ to a $q\notin U(V_{\phi})$.
For the second question, clearly $V_\phi\subseteq W_\phi\subseteq \overline{V_\phi}$, so $W_\phi = \overline{V_\phi}$. Thus, $W_\phi=V_\phi$ iff $\Phi$ satisfies the condition in the Proposition above.
