It is known that if $\mathcal{U}$ be a uniformity on $X$, then every entourage $N\in\mathcal{U}$ is an open set of diagonal $\Delta_X$, but the converse of it, is not true.
For instance, Consider $\mathbb{R}$ with usual metric $d$. For every $\epsilon>0$, let $$U^d_\epsilon:=\left\{(x, y)\in \mathbb{R}^2 : d(x, y)<\epsilon\right\}$$ Then the collection $$ \mathcal{U}_d=\left\{E\subseteq \mathbb{R}^2 : U_\epsilon^d\subseteq E, \text{ for some } \epsilon>0\right\}$$ is a uniformity on $\mathbb{R}$.
In this example, every element of $\mathcal{U}_d$ is a neighborhood of $\Delta_\mathbb{R}$ in $\mathbb{R}^2$ but $\left\{(x, y) : |x-y|<\frac{1}{1+|y|}\right\}$ is a neighborhood of $\Delta_\mathbb{R}$ but not a member of $\mathcal{U}_d$.
In my research, $(X, d)$ is a locally compact, $N=\{(x, y): d(x, y)<\delta \}$ and $\mathcal{U}$ is an uniformity of $X$. Can I say that there is $D\in\mathcal{U}$ with $D\subseteq N$?
No, an entourage need not be an open neighbourhood of $\Delta_X$, but it is a neighbourhood of $\Delta_X$. E.g. $\{(x,y): d(x,y) \le \varepsilon\}$ is not open in $X \times X$ in general but it is a neighbourhood of the diagonal as it contains $\{(x,y): d(x,y) < \varepsilon\}$ as a subset, which happens to be open. But an (even open) neighbourhood of $\Delta_X$ need not be an entourage, as your example indeed shows.
As to your final question: the answer is no, I think. Use another equivalent metric on $\mathbb{R}$ and its induced uniformity, to see this: if this property were true it would hold for $\mathbb{R}$ (locally compact) and then it would imply that all metrics on the reals would be uniformly equivalent, which they're not.