Let me first explain how one arrives at $(1)$. The summation in $(0)$ is over all pairs $\langle k,\ell\rangle$ of non-negative integers. For non-negative integers $k$ and $\ell$ let
$$a(k,\ell)=g_kh_\ell\delta\big(t-(k+\ell)T\big)\;;$$
then we can abbreviate $(0)$ as
$$\sum_{k=0}^\infty\sum_{\ell=0}^\infty a(k,\ell)\;.\tag{3}$$
Think of the numbers $a(k,\ell)$ as being laid out in an infinite array with rows and columns each numbered $0,1,2,\dots$; $a(k,\ell)$ is the number in row $k$, column $\ell$. $(3)$ calculates the sum of the entries in this array by first summing each row: the sum of the entries in row $k$ is $\sum_{\ell=0}^\infty a(k,\ell)$. It then adds up the row sums to get the grand total. (I’m assuming that in context there is information to guarantee convergence of these infinite sums.)
But assuming convergence we can sum the entries in other orders. In particular, for each non-negative integer $n$ we can sum along the anti-diagonal $k+\ell=n$, and we can then add up these anti-diagonal sums. (The diagonal of this infinite array consists of the entries $a(k,k)$; the set of entries $a(k,\ell)$ with $k+\ell=n$ for some fixed $n$ is perpendicular to the diagonal, i.e., an anti-diagonal.) When $n=0$, $\langle k,\ell\rangle$ can only be $\langle 0,0\rangle$; when $n=1$, $\langle k,\ell\rangle$ can be $\langle 0,1\rangle$ or $\langle 1,0\rangle$; when $n=2$, $\langle k,\ell\rangle$ can be $\langle 0,2\rangle$, $\langle 1,1\rangle$, or $\langle 2,0\rangle$; and so on. As you can see, for each value of $n\ge 0$, the possible values of $k$ are $0,1,\dots,n$, and of course in each case $\ell=n-k$.
If $n=k+\ell$, we have $a(k,\ell)=a(k,n-k)$, and the sum of the entries along the $n$-anti-diagonal is $\sum_{k=0}^n a(k,n-k)$. The grand total is the sum of these anti-diagonal sums, which is
$$\sum_{n=0}^\infty\sum_{k=0}^n a(k,n-k)\;.\tag{4}$$
Now
$$a(k,\ell)=a(k,n-k)=g_kh_{n-k}\delta(t-nT)\;,$$
so when we expand $(4)$ we get
$$\sum_{n=0}^\infty\sum_{k=0}^ng_kh_{n-k}\delta(t-nT)\;.\tag{5}$$
The nice thing about this arrangement is that the expression $\delta(t-nT)$ depends only on $n$, not on $k$, so we can factor it out of the inner summation to get
$$\sum_{n=0}^\infty\left(\sum_{k=0}^ng_kh_{n-k}\right)\delta(t-nT)\;,\tag{1}$$
which we could just as well write
$$\sum_{n=0}^\infty\delta(t-nT)\sum_{k=0}^ng_kh_{n-k}\;.$$
Your $(2)$ is not wrong, but it will be easier to work with if you observe that $n-k\ge 0$ if and only if $n\ge k$, so that it can be rewritten
$$\sum_{k=0}^\infty\sum_{n=k}^\infty g_kh_{n-k}\delta(t-nT)\;.$$
Getting the clutter out of the way again, we can write this as
$$\sum_{k=0}^\infty\sum_{n=k}^\infty a(k,n-k)\;.\tag{6}$$
Turning $(6)$ into $(4)$ is a matter of reversing the order of summation. What are the possible values of $n$ in $(6)$? They depend on $k$, of course, since we always have $n\ge k$, but since $k$ can be as small as $0$, so can $n$. Thus, $n$ ranges over the non-negative integers, and when we reverse the order of summation, the outer summation must be $\sum_{n=0}^\infty$. For each value of $n$, what are the possible values of $k$? Again, we know that $k\le n$, and $k$ can be as small as $0$, so for each value of $n$ we must let $k$ run from $0$ through $n$. In other words, when we reverse the order of summation and revise the limits appropriately, $(6)$ becomes
$$\sum_{n=0}^\infty\sum_{k=0}^n a(k,n-k)\;,$$
which is exactly $(4)$, as we wished.
