Maybe this will help you think about the coefficients. We have $n$ boys and $m$ girls, and we want to form a committee of $l$ people from these $n+m$ people. Clearly there are $\binom{n+m}{l}$ ways to form the committee.
Let's count the number of committees another way. We could have $0$ boys and $l$ girls. Such a committee can be formed in $\binom{n}{0}\binom{m}{l}$ ways.
Or we could have $1$ boy and $l-1$ girls. Such a committee can be formed in $\binom{n}{1}\binom{m}{l-1}$ ways.
Or we could have $2$ boys and $l-2$ girls. Such a committee can be formed in $\binom{n}{2}\binom{m}{l-2}$ ways.
Continue. The total number of committees is
$$\sum_{k=0}^l \binom{n}{k}\binom{m}{l-k}.\tag{$1$}$$
But we already saw that the number of committees is $\binom{n+m}{l}$.
Note: It is possible that for example there is a total of $3$ boys and $24$ girls, and we want to form a committee of $7$ people. Then the formula appears to break down. But it doesn't if we agree that $\binom{a}{b}=0$ if $b\gt a$.
To apply the reasoning to $(1+x)^n(1+x)^m$, you might first look at $(1+x)^n(1+y)^m$, and set $y=x$ at the end. So expand both, multiply. For fixed $l$, gather together terms that have a combined total of $l$ $x$'s (boys) and/or $y$'s. The total number will be given by Formula $(1)$.
Call $n+1$-th element $e_{n+1}$. What shall the equivalence class of $e_{n+1}$ be?
Maybe $e_{n+1}$ will be in an equivalence class all by itself. Then by definition the other $n$ objects can be divided into equivalence classes in $q_n$ ways. For consistency with later stuff we say that there are $\binom{n}{0}q_n$ ways to divide our set of $n+1$ elements into equivalence classes so that $e_{n+1}$ is a loner.
Maybe $e_{n+1}$ will have only $1$ other object in its equivalence class, there will be only one other object in the family of $e_{n+1}$. Which object? It can be chosen in $\binom{n}{1}$ ways. For every one of these ways, the remaining $n-1$ objects can be divided into equivalence classes in $q_{n-1}$ ways, giving a total of $\binom{n}{1}q_{n-1}$ divisions of our set for which the equivalence class of $e_{n+1}$ contains only one object other than $e_{n+1}$.
Maybe $e_{n+1}$ will have $2$ other objects in its equivalence class. Which $2$ objects? They can be chosen in $\binom{n}{2}$ ways. For every one of these ways, the remaining $n-2$ objects can be divided into equivalence classes in $q_{n-2}$ ways, giving a total of $\binom{n}{2}q_{n-2}$ of this type.
Maybe $e_{n+1}$ will have $3$ other objects in its equivalence class. These objects can be chosen in $\binom{n}{3}$ ways. For every one of these ways, the remaining $n-3$ can be "split up" in $q_{n-3}$ ways.
Continue. The expression of the problem basically summarizes the analysis given above.
To use the notation of the formula, the equivalence class of $e_{n+1}$ will contain $k$ objects other than $e_{n+1}$, where $k$ is any of $0$, $1$, $2$, and so on up to $n$. The number of ways to have the equivalence class of $e_{n+1}$ have $k$ objects other than $e_{n+1}$ is $\binom{n}{k}$ (choose the rest of the family) times $q_{n-k}$ (split the others into families). Now sum over all $k$.
Best Answer
Let $k=n+1$, so that $n=k-1$, and substitute into $$\sum_{n\ge 0}\binom{n+2}2z^{n+1}$$ to get $$\sum_{k-1\ge 0}\binom{(k-1)+2}2z^k=\sum_{k\ge 1}\binom{k+1}2z^k\;;$$ then simply rename $k$ back to $n$ to get
$$\sum_{n\ge 1}\binom{n+1}2z^n\;.$$
(Note that it’s now $n+1$ in the binomial coefficient, not $n+2$.) Finally,
$$\binom{n+1}2=\frac{(n+1)!}{2!\big((n+1)-2\big)!}=\frac{(n+1)!}{2(n-1)!}=\frac{n(n+1)}2\;.$$