Contrary to the suggestions of the other users, I believe this can be proved (the counterexample provided does not address the OP's problem as OP clarified it in the comments).
Let the original graph be $G$, the new graph be $G^\prime$, the vertices on the left be $V_1$, the vertices on the right be $V_2$, and the source and sink be $s$ and $t$ respectively.
$|\text{min cut}| \leq |\text{min vertex cover}|$:
Let $C_1 \cup C_2$, where $C_1 \subseteq V_1, C_2 \subseteq V_2$, be the smallest vertex cover. Consider the following cut:
$S = \{s\} \cup (V_1 \setminus C_1 ) \cup C_2$
Then
$\bar{S} = \{t\} \cup C_1 \cup (V_2 \setminus C_2)$
Note that none of the arcs from $V_1$ to $V_2$ are present in the cut. We can see this as follows. If an arc starts from a vertex in $V_1$ inside the vertex cover, then it is not in $S$ in the first place. If an arc starts from a vertex in $V_1$ not in the vertex cover, then it must end in a vertex in $V_2$ in the cover (otherwise we would not have a vertex cover). Therefore, the arc must not be in the cut, since it connects two vertices within $S$.
As a result, the only arcs in the cut are those from $\{s\}$ to $C_1$ and from $C_2$ to $\{t\}$. The size of this cut is $|C_1 \cup C_2|$, or the size of the minimum vertex cover. Therefore, the size of the min cut is at most the size of the min vertex cover.
$|\text{min cut}| \geq |\text{min vertex cover}|$:
Let the minimum cut be $(S,\bar{S})$. We express the cut as follows:
$S = \{s\} \cup (V_1 \setminus C_1 ) \cup C_2$
$\bar{S} = \{t\} \cup C_1 \cup (V_2 \setminus C_2)$
where $C_1 \subseteq V_1$ and $C_2 \subseteq V_2$. Clearly, the arcs inside this cut consist of all the arcs from $\{s\}$ to $C_1$, from $V_2 \cap C_2$ to $\{t\}$, and from $V_1 \setminus C_1$ to $V_2 \setminus C_2$.
Consider the set $C_1 \cup C_2$. We may expand this set to a vertex cover by adding all the vertices of $V_1 \setminus C_1$ for which an arc from it to $V_2 \setminus C_2$ exists. Obviously the size of this set is the same as the size of the min cut. So the size of the minimum vertex cover is at most the size of the minimum cut and we are done.