For simplicity write all the numbers in base $4$. Fix an arbitrary ordering of the edges of the graph $G$ and let $e_i$ be the $i$-th edge ($i= 1, \dots, |E(G)|$).
For each vertex $v \in V(G)$ define an integer $n_v$ of $|E(G)| + 1$ digits as follows:
- the $i$-th least significant digit of $n_v$ is $1$ if $e_i$ is incident to $v$ and $0$ otherwise;
- the most significant digit of $n_v$ is $1$.
For each edge $e_i \in E(G)$ let $m_{e_i} = 4^{i-1}$ (i.e., the $i$-th least significant digit of $m_e$ is $1$ and all the other digits are $0$).
The instance of subset sum is obtained by choosing $S = \{ n_v : v \in V(G) \} \cup \{m_e : e \in E(G) \}$ and $t = k 4^{|E(G)|} + 2\sum_{i=0}^{|E(G)|-1} 4^i$.
In other words, the most significant digits of $t$ encode the number $k$ in base $4$, while the $|E(G)|$ least significant digits of $t$ are all equal to $2$.
If there is a vertex cover $V' \subseteq V(G)$ of $G$ of size $k$,
let $E'$ be the set of edges $(u,v) \in E(G)$ such that exactly one of $u$ and $v$ is in $V'$.
Then, $A = \{ n_v : v \in V' \} \cup \{ m_e : e \in E' \}$ is a subset of $S$ that sums to $t$. Indeed:
$$
\sum_{x \in A} x = \sum_{v \in V'} n_v + \sum_{e \in E'} m_v
= k 4^{|E(G)|} + 2 \sum_{e_i \in E \setminus E'} 4^{i-1} +
2 \sum_{e_i \in E'} 4^{i-1}
= k 4^{|E(G)|} + 2 \sum_{e_i \in E} 4^{i-1} = t.
$$
If there is a subset $A$ of $S$ that sums to $t$, then $A$ contains exactly $k$ numbers from $\{ n_v : v \in V(G) \}$ (as otherwise the most significant digits of $\sum_{x \in A} x$ would not match those of $t$). Moreover, $V' = \{v_i : n_i \in A \}$ is a vertex cover for $G$.
To see this, pick any $e_i = (u,v) \in E(G)$ and notice the $i$-th least significant digit of $t$ is $2$.
Since, for $j=1,\dots, |E(G)|$, there are exactly $3$ numbers in $S$ whose $j$-th least significant digit is non-zero (namely $n_{u'}$, $n_{v'}$, and $m_{e_j}$, where $e_j = (u', v')$),
we have that in the summation $\sum_{x \in A} x$ no carry occurs among the $|E(G)|$ least significant digits. We conclude that $A$ contains exactly two numbers from $\{n_u, n_v, m_e\}$, implying that $u \in V'$ or $v \in V'$ (possibly both).
In your example graph, for $k=3$ you get the following (bold rows denote the elements of $A$ when $V'=\{v_1, v_3, v_4\}$):
$$
\begin{array}{|c|c|c|c|c|}
\hline
& & e_1 & e_2 & e_3 & e_4 \\ \hline
\pmb{n_{v_1}} & \pmb{1} & \pmb{1}& \pmb{0}& \pmb{0}& \pmb{1}\\ \hline
n_{v_2} & 1 & 1& 1& 0& 0\\ \hline
\pmb{n_{v_3}} & \pmb{1} & \pmb{0} & \pmb{1} & \pmb{1} & \pmb{0}\\ \hline
\pmb{n_{v_4}} & \pmb{1} & \pmb{0} & \pmb{0} & \pmb{1} & \pmb{1}\\ \hline
\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|}
\hline
& & e_1 & e_2 & e_3 & e_4 \\ \hline
\pmb{m_{e_1}} & \pmb{0} & \pmb{1} & \pmb{0} & \pmb{0} & \pmb{0} \\ \hline
\pmb{m_{e_2}} & \pmb{0} & \pmb{0}& \pmb{1}& \pmb{0}& \pmb{0}\\ \hline
m_{e_3} & 0 & 0& 0& 1& 0\\ \hline
m_{e_4} & 0 & 0& 0& 0& 1\\ \hline
\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|}
\hline
\phantom{m_{e_4}}& & e_1 & e_2 & e_3 & e_4 \\ \hline
t & 3 & 2& 2& 2& 2\\ \hline
\end{array}
$$
