What matters is to express the Nash equilibrium conditions in terms of the winning probabilities $x_i$ and then find a concave “potential” whose KKT conditions reproduce those equilibrium conditions. In your contest, $x_i=e_i/E$ with $E=\sum_j e_j$. Player $i$’s payoff is $u_i(e)=v_i x_i-e_i$. Holding rivals’ efforts fixed, the marginal effect of $e_i$ on $x_i$ is $$ \frac{\partial x_i}{\partial e_i}=\frac{E-e_i}{E^2}=\frac{1-x_i}{E}. $$ Hence the first-order condition for any player with $e_i>0$ is $$ \frac{\partial u_i}{\partial e_i}=v_i\,\frac{\partial x_i}{\partial e_i}-1 =\frac{v_i(1-x_i)}{E}-1=0, $$ i.e. $$ v_i(1-x_i)=E \quad\text{if } e_i>0,\qquad v_i(1-x_i)\le E \quad\text{if } e_i=0. $$ Together with the feasibility $\sum_i x_i\le 1$ (with equality when $E>0$), these inequalities exactly characterize the Nash equilibrium in $x$-space.
Now take the convex program $$ \max_{x\ge 0,\;\sum x_i\le 1}\; \sum_i \hat u_i(x_i), \qquad\text{with }\;\hat u_i'(x_i)=v_i(1-x_i). $$ If $\lambda$ is the multiplier on $\sum_i x_i\le 1$ and $\mu_i$ the multipliers on $x_i\ge 0$, the KKT conditions for an interior $x_i>0$ are $$ \hat u_i'(x_i)-\lambda=0 \;\Longleftrightarrow\; v_i(1-x_i)=\lambda, $$ and for a boundary $x_i=0$: $v_i(1-x_i)\le \lambda$. Comparing with the Nash conditions above shows that the Lagrange multiplier $\lambda$ plays exactly the role of the aggregate effort $E$, so any solution of the program coincides with the contest equilibrium. Because $\hat u_i''(x_i)=-v_i<0$, the objective is strictly concave and the solution is unique.
Integrating $\hat u_i'(x_i)=v_i(1-x_i)$ gives $$ \hat u_i(x_i)=\int_0^{x_i} v_i(1-z)\,dz=v_i\Big(x_i-\tfrac12 x_i^2\Big) =v_i x_i\Big(1-\tfrac12 x_i\Big), $$ which is the formula you quoted (up to an irrelevant constant). This is where the $\hat u_i$ comes from: it is a potential whose gradient reproduces each player’s marginal payoff once the effort externality is represented by the common constraint $\sum x_i\le 1$.
Your alternative expression $$ \hat u_i(x_i)=(1-x_i)u_i(x_i)+x_i\Big(\frac{1}{x_i}\int_0^{x_i} u_i(z)\,dz\Big) =(1-x_i)u_i(x_i)+\int_0^{x_i} u_i(z)\,dz $$ is just a convenient antiderivative identity. Differentiating yields $$ \frac{d}{dx_i}\hat u_i(x_i)=(1-x_i)\,u_i'(x_i), $$ so whenever the “gross” utility in $x$-space is $u_i(x_i)=v_i x_i$ (the prize term), you get $\hat u_i'(x_i)=(1-x_i)v_i$ and hence the same $\hat u_i=v_i(x_i-\tfrac12 x_i^2)$. Note that when you rewrote $u_i$ as $v_i x_i-\frac{x_i}{\sum_j e_j}$ you dropped a factor: since $e_i=x_i\sum_j e_j$, the cost term is $-e_i=-x_i\sum_j e_j$, not $-x_i/\sum_j e_j$. The cost is absorbed in the optimization via the shared constraint and its multiplier $\lambda=E$, which is precisely why maximizing $\sum_i \hat u_i(x_i)$ delivers the equilibrium $x$-vector.
