Finally I found how to do it. I post it, if someone is interested.
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}-\varkappa^2/4}e^{\varkappa^2/4}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4}\underbrace{\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-(\tilde{x}-\varkappa/2)^2}H_l(\tilde{x})\;\mathrm{d}\tilde{x}}_I
\end{align}
If we pose $x = \tilde{x}-\frac{\varkappa}{2}$ in this expression, the integral $I$ becomes
\begin{equation*}
I = \int_{-\infty}^{+\infty}H_n(x+\varkappa/2)e^{-x^2}H_l(x+\varkappa/2)\;\mathrm{d}x
\end{equation*}
We know that
\begin{equation*}
H_n(x+a) = \sum_{p=0}^n \frac{n!}{(n-p)!p!}(2a)^{n-p}H_p(x)
\end{equation*}
Hence, the integral $I$ becomes
\begin{align*}
I &= \int_{-\infty}^{+\infty} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\varkappa^{n-p}H_p(x) e^{-x^2} \sum_{q=0}^l \frac{l!}{(l-q)!q!}\varkappa^{l-q}H_q(x)\;\mathrm{d}x \\
&= \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\varkappa^{n-p}\frac{l!}{(l-q)!q!}\varkappa^{l-q}\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x \\
\end{align*}
The Hermite polynomials are orthogonal in the range $(-\infty,\infty)$ with respect to the weighting function $e^{-x^2}$ and satisfy
\begin{alignat*}{2}
&&&\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x = \sqrt{\pi}2^pp!\;\delta_{pq} \\
&\Rightarrow\quad&& I = \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\frac{l!}{(l-q)!q!}\varkappa^{n+l-p-q}\cdot \sqrt{\pi}2^pp!\;\delta_{pq}
\end{alignat*}
As this integral is nil if we have not $p=q$, we can replace the two sums by only one that goes from 0 to $\min(n,l)$. Let us say that $n<l$. Hence, the full expression for the $D$-matrix is
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!p!}2^pp!\sqrt{\pi}\;\varkappa^{n+l-2p} \notag\\
&= \frac{\varkappa^{n+l}}{\sqrt{2^nn!}\sqrt{2^ll!}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!}2^p\;\varkappa^{-2p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{n+l}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{-p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{n-p}
\end{align}
Associated Laguerre polynomials $L_a^b(x)$ are given by
\begin{equation*}
L_a^b(x) = \sum_{k=0}^{a}(-1)^k \frac{(a+b)!}{(a-k)!(b+k)!k!}x^k
\end{equation*}
It suggests us to transform the expression of the $D$-matrix by posing $k=n-p$. Hence, we have
\begin{align}
D_{ln}(\varkappa) &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=n}^0 \frac{(l)!}{(n-(n-k))!(l-(n-k))!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{n-(n-k)} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n \frac{l!}{k!(l-n+k)!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n (-1)^k\frac{([l-n]+n)!}{(n-k)!([l-n]+k)!k!}\left(-\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4}L_n^{l-n}\left(-\frac{\varkappa^2}{2}\right)
\end{align}
It should be remembered that we had supposed that $n<l$. But that could be otherwise. In order to be general, $n_<$ and $n_>$ will be defined as $n_<=\min{(n,l)}$ and $n_>=\max{(n,l)}$ and $l-n=|l-n|$. We then have
\begin{equation}
D_{ln}(\varkappa) = \sqrt{\frac{n_<!}{n_>!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{|l-n|}L_{n_<}^{|l-n|}\left(-\frac{\varkappa^2}{2}\right)e^{\varkappa^2/4}
\end{equation}
Best Answer
It's not easy to derive one definition from the other, but here's one possible way: the Hermite polynomials can also be expressed in terms of their generating function, that is a function $g(t,x)$ such that $$ g(t,x) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ It follows that $$ H_n(x) = \left.\frac{\partial^n}{\partial t^n}g(t,x)\right|_{t=0} $$ Now, if we take the first definition $$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}, $$ we see that it looks quite similar. We need to find a way to change the derivatives to $x$ into derivatives to $t$, and we can do that by noting that $$ \frac{\partial}{\partial t}e^{-(t-x)^2} = -\frac{\partial}{\partial x}e^{-(t-x)^2} $$ Thanks to this trick, the function $g(t,x)$ follows immediately: $$ g(t,x) = e^{x^2}e^{-(t-x)^2} = e^{2xt-t^2}. $$ This allows us to derive yet another form for the Hermite polynomials, by equating the terms in $$ e^{2xt-t^2} = \sum_{n=0}^\infty\frac{(2xt-t^2)^n}{n!} = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ Let us examine the $(n-k)$th term in the first series: $$ (2x-t)^{n-k}\frac{t^{n-k}}{(n-k)!} = \sum_{l=0}^{n-k}\binom{n-k}{l}(2x)^{n-k-l}(-1)^l\frac{t^{n-k+l}}{(n-k)!}. $$ The $l=k$ term in this sum contains $t^n$: $$ (-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ provided that $2k\leqslant n$. In other words, all the terms that contribute to the $t^n$ term are $$ H_n(x)\frac{t^n}{n!} = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ so that we've found another expression for $H_n(x)$: $$ H_n(x) = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{n!}{k!\,(n-2k)!}. $$ We can write this as $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}\frac{(n-k)!}{(n-2k)!}(-1)^k(2x)^{n-2k}, $$ which is $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}(-1)^k\frac{d^k}{dx^k}(2x)^{n-k}, $$ and this is a more explicit notation of the other definition $$ \left(2x - \frac{d}{dx}\right)^n (1) $$ In principle, you can follow the steps in the reverse order to derive the first definition from the second, but it's certainly not intuitive. Alternatively, one might use the recurrence relations $$ H_{n+1}(x) = 2xH_n(x) - 2n H_{n-1}(x) $$ $$ H'_n(x) = 2nH_{n-1}(x). $$ It's easy to verify that both definitions satisfy these relations.