Here is an extended hint for the first conjectured formula.

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Write the sum in the exponential above as 

\begin{align*}
\sum_{n>0} \frac{q^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w^j\Big)
\end{align*}

and rewrite the inner sums as 

\begin{align*}
\sum_{j>0}[z^{n+j}]\{f(z)^n\} w^j =\frac{1}{w^n}\Big(f(w)^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w^j\Big) -[z^n] \{f(z)^n\}
\end{align*}

Note that 
\begin{align*}
\sum_{n>0} \frac{q^n}{n}[z^n]\{f(z)^n\}=\log\big(f(w)\big)
\end{align*}
and that the remaining sum can be written as
\begin{align*}
\sum_{n>0} \frac{1}{n} [u^{n-1}]\Big\{\frac{1}{1-u} \big(1-\frac{f(uw)^n}{f(w)^n}\big)\Big\}
\end{align*}

Summing up and evaluating the residue now gives the value $-\log\big(1-q\,f^\prime(w)\big)$  
for this sum, as follows:

\begin{align*}
&\sum_{n>0} \frac{1}{n} [u^{n-1}]\Big\{\frac{1}{1-u} \big(1-\frac{f(uw)^n}{f(w)^n}\big)\Big\}\\
  &= [u^{-1}]\Big\{\frac{1}{1-u}\big(-\log(1-\frac{1}{u}) +\log(1-\frac{f(uw)}{uf(w)})\big)\Big\}\\
  &= [u^{-1}]\Big\{\frac{1}{1-u}\Big(\log\big(\frac{1-\frac{f(uw)}{uf(w)}}{1-\frac{1}{u}}\big)\Big)\Big\}\\
\end{align*}
Now substitute $u=z+1$ and use the residue composition rule (for formal Laurent series, see e.g. Goulden/Jackson, *Combinatorial Enumeration*, Thm 1.2.2).
\begin{align*}
  &=[z^{-1}]\Big\{-\frac{1}{z}\log\Big(\frac{1+z-\frac{f((1+z)w)}{f(w)}}{z}\Big)\Big\}\\
  &=[z^{0}]\Big\{-\log\Big(\frac{1+z-\frac{f((1+z)w)}{f(w)}}{z}\Big)\Big\}\\
  &=-\log\big(1-qf^\prime(w)\big)
\end{align*}

Finally, note that 
\begin{align*}
\frac{1}{f(w)}\frac{1}{1-q\,f^\prime(w)}= \frac{q^2 w^\prime}{w^2}\;\;.
\end{align*}