Let $V$ a finite dimensional vector space with $dim\;V=n$ and $V'$ be its dual space over a field $\Bbb F$.
This can be proved in two ways:
$\mathbf {Proof \;1}:$ Let $B_S=\{s_1,...,s_k\}$ be a basis for the subspace $S$. So $dim\;S=k$. Extend $B_S$ to a basis of $V$ say $B_V=\{s_1,...,s_k,s_{k+1},...,s_n\}$. Let $B^{'}_V=\{f_1,...,f_n\}$ be the dual basis to $B_V$ where $f_i(s_j)=\delta_{i,j}$, the Kronecker Delta.
My Claim is that $\{f_{k+1},...,f_n\}$ is the basis of $N=S^0$. Indeed let $f\in V'$ then $f=a_1f_1+...+a_nf_n$ for some $a_1,...,a_n \in \Bbb F$. Now $f(s_j)=a_jf_j(s_j)=a_j$. So $f=f(s_1)f_1+...+f(s_n)f_n$. If $i\in \{k+1,...,n\}$ and $j \in \{1,...,k\}$ then $f_i(s_j)=0$. So if $s\in S$ then $s=b_1s_1+...+b_ns_k$ for some $b_1,....,b_k\in \Bbb F $. Then $f_i(s)=0\; \forall i \in \{k+1,...,n\}$. Hence $f_{k+1},...,f_n \in S^0$. Then $ span \{f_{k+1},...,f_n \}\subseteq S^0$ since $S^0$ is a subspace. Now suppose $f\in S^0$ then $f(s_1)=...=f(s_k)=0$ since $s_1,...,s_k \in S$. Then writing $f=f(s_1)f_1+...+f(s_n)f_n$, the first $k$ terms are zero i.e $f=f(s_{k+1})f_{k+1}+...+f(s_n)f_n$. So $f\in span\{f_{k+1},...,f_n\}$. Hence $S^0\subseteq span\{f_{k+1},...,f_n\}$. Hence $S^0=span\{f_{k+1},...,f_n\}$, and of course $\{f_{k+1},...,f_n\}$ is linearly independent being a subset of a set of basis vectors.So $dim\;S^0=n-k=dim\;V-dim\;S$.So $dim\;V=dim\;S + dim\;S^0$ . $\lhd$
$\mathbf {Proof \;2}:$ Now suppose $i:S\rightarrow V$ be the natural inclusion map i.e $i(s)=s,\; \forall s \in S$. Consider its dual map $i':V'\rightarrow S'$ defined as follows $i'(\phi)=\phi\,\circ i$. Thus $i'$ is a linear map from $V'$ to $S'$. Then $dim\,range(i')+dim\,null(i')=dim\,V'=dim\,V$. Now my first claim is,
$\it{Claim \;1}:$ $null(i')=S^0.$
$\it{Pf}:$ If $\phi \in null(i')\Leftrightarrow i'(\phi)=0\Leftrightarrow \phi\,\circ i=0\Leftrightarrow(\phi\,\circ i)(s)=0,\,\forall s\in S\Leftrightarrow\phi(i(s))=0,\,\forall s\in S$ $\Leftrightarrow\phi(s)=0,\, \forall s\in S\Leftrightarrow \phi \in S^0\,\bullet$
Now we have $dim\,range(i')+dim\,S^0=dim\,V$. This means that $dim\,range(i')$ must be equal to $dim\,S=dim\,S'$. But $range\,(i')\subseteq S'$. So it means that $i'$ must be surjective i.e $range(i')=S'$. Indeed,
$\it{Claim\;2}:$ $range(i')=S'$.
$\it{Pf}:$ Let $\phi \in S'$. So $\phi:S\rightarrow \Bbb F$. We extend $\phi $ to a functional $\psi :V\rightarrow \Bbb F$ such that $\psi(s)=\phi(s),\,\forall s\in S$. So $i'(\psi)=\psi\,\circ i$. But $(\psi\, \circ i)(s)=\psi(i(s))=\psi(s)=\phi(s),\,\forall s \in S$. Hence $\psi\, \circ i = \phi \Leftrightarrow i'(\psi)=\phi$. So $S'=range(i')\,\bullet$.
Finally we have $dim\,S'+dim\,S^0=dim\,V\Rightarrow dim\,S+dim\,S^0=dim\,V.\,\lhd$
For a subspace $T\subseteq V'$ define $T^0 := \{v\in V\,|\, \varphi(v) = 0\text{ for all $\varphi\in T$}\}$ (which is the same as your definition under the identification $V = V''$). In your case, we have $\Gamma^0 = S$.
Then we have $\dim V = \dim U + \dim U^0$ for each subspace $U\subseteq V$: Take a basis $u_1,\dotsc,u_d$ of $U$ and complete it to a basis $u_1,\dotsc,u_n$ of $V$. Denoting by $u'_1,\dotsc,u'_n$ the dual basis, it is easy to see that $U^0$ is spanned by $u'_{d+1},\dotsc,u'_n$, which proves the claim.
Likewise, we have $\dim V' = \dim T + \dim T^0$ for any subspace $T\subseteq V'$. More precisely, let $\varphi_1,\dotsc,\varphi_d$ be a basis of $T$ and complete it to a basis $\varphi_1,\dotsc,\varphi_n$ of $V'$. Then it follows directly from the definitions that $T^0 = \bigcap_{i=1}^d \ker\varphi_i$. It remains to check that $\dim \bigcap_{i=1}^d \ker \varphi_i = n-d$. Indeed, putting $W_j = \bigcap_{i=1}^j \ker\varphi_i$, for $j=1,\dotsc,d$, we obtain an increasing chain
$$
0 = W_n \subset W_{n-1} \subset\dotsb \subset W_1 \subset V.
$$
Since obviously $\dim W_j \ge \dim W_{j+1} \ge \dim W_j-1$, for all $j$, and $\dim W_1 = n -1$, we must in fact have $\dim W_j = n-j$. In particular,
$$
\dim T^0 = \dim \bigcap_{i=1}^d\ker\varphi_i = \dim W_d = n-d = \dim V' - \dim T.
$$
Now, it follows that $T^{00} = T$ for $T\subseteq V'$: $T\subseteq T^{00}$ is clear and both have the same dimension, because of
$$\dim T = \dim V' - \dim T^0 = \dim V - \dim T^0 = \dim T^{00}.$$
With this, we conclude $\Gamma = \Gamma^{00} = S^0$ (using $\Gamma^0 = S$ for the last equality).
Best Answer
We have
$$\varphi_k(v_j) = \begin{cases} 1 & \text{if } k = j\\ 0 & \text{if } k \ne j\text{} \end{cases}$$
So every element of $\{\varphi_{m+1}, \dots, \varphi_{n}\}$ is zero on each basis element of $U$, hence $\{\varphi_{m+1} , \dots,\varphi_n \} \subseteq U^\circ$.