Residue identity for function composition: $ \operatorname{Res}(f; h(a)) = \operatorname{Res}((f∘h)h'; a)$ With $a \in \mathbb{C}$, let $h: D_1 \to D_2$ and $f: D_2\setminus{\{h(a)\}} \to \mathbb{C}$ be analytic functions. Also require $h'(a) \neq 0$.

Claim: $$ \operatorname{Res}(f; h(a)) =  \operatorname{Res}((f \circ h)h'; a)$$

Attempt
For $\rho$ small enough, we have:
$$ \operatorname{Res}(f; h(a)) = \frac{1}{2\pi i}\int_{|u-h(a)|=\rho} f(u)\ \mathrm{d}u$$
and
\begin{align}\operatorname{Res}((f \circ h)h', a)& = \frac{1}{2\pi i} \int_{|u-a|=\rho} (f \circ h) h'\ \mathrm{d}u = \frac{1}{2\pi i} \int_0^{2\pi} f(h(a + \rho e^{it})) h'(a + \rho e^{it})\rho e^{it} i\ \mathrm{d}t\\&= \frac{1}{2\pi i} \int_\alpha f(u)\ \mathrm{d}u
\end{align}
where $\alpha: [0, 2\pi] \to \mathbb{C}, \alpha(t) = h(a+\rho e^{it})$.
We see that the curve of the LHS defined by $|u-h(a)|=\rho$ (i.e. $h(a) + \rho e^{it}$) is "approximated" by the curve defined by $\alpha$ for $\rho \to 0$ because analytic functions approximately retain circles (see here).
(Note that we already used $h'(a) \neq 0$: Writing the residue for the RHS was only possible because $(f \circ h)$ is analytic on $U_\rho(a)\setminus{\{a\}}$. Since $h'(a) \neq 0$, $h$ is injective on a small enough neighborhood around $a$.)
Capturing the approximation
I tried to generalize the claim above by introducing the following 

Lemma: If $f: D \to \mathbb{C}$ is a continuous function and $(\alpha_n)_n, (\beta_n)_n \subseteq D$ series of piecewise smooth curves $\alpha_n, \beta_n: [a, b] \to D$ such that $(f \circ \alpha_n) \alpha_n' - (f \circ \beta_n) \beta_n' \to 0$ uniformly, then:
  If for a fixed $z \in \mathbb{C}$ the equation holds: $z = \int_{\alpha_n} f(u)\ \mathrm{d}u$ for all $n \ge 0$
  then $z = \int_{\beta_n} f(u)\ \mathrm{d}u$.

Proof: For every $\varepsilon > 0$ we have $|z - \int_{\beta_n} f(u)\ \mathrm{d}u| = \left|\int_a^b ((f \circ \alpha_n) \alpha_n' - (f \circ \beta_n) \beta_n')(t)\ \mathrm{d}t\right| \le (b - a) \sup_{t \in [a, b]} \left|...\right| \le \varepsilon$ for a chosen $n$ big enough.
Using the lemma
Choose $\alpha_n(t) := h(a + \rho_n e^{it})$ (from the RHS), $\beta_n(t) := h(a) + \rho_n e^{it}$ (from the LHS).
We have
$$|\alpha_n(t) - \beta_n(t)| = \left|f(h(a + \rho_n e^{it})) h'(a + \rho_n e^{it})\rho_n e^{it} i - f(h(a) + \rho_n e^{it}) \rho_n e^{it} i\right|
= |\rho_n| \cdot |*|$$
where
$$|*| = \left|f(h(a + \rho_n e^{it})) h'(a + \rho_n e^{it}) - f(h(a) + \rho_n e^{it})\right|.$$
If $|*|$ was bounded, we would be able to apply the theorem. This is the place where I am currently stuck. I tried expressing $f$ with its Laurent series around $h(a)$, but that did not lead to easier-looking equations.
 A: Let $U \subset D_1$ be a disk with center $a$ in which $h$ is injective.
Then $V := h(U) \subset D_2$ is a simply-connected neighbourhood of $h(a)$.
For $\rho > 0$ small enough, let
$$
 \gamma: [0, 2 \pi] \to U, \gamma(t) = a + \rho e^{it}
$$
and $\alpha = h \circ \gamma$. Then – as you already computed –
$$
 \operatorname{Res}((f \circ h)h', a) = \frac{1}{2 \pi i}\int_\gamma f(h(z))h'(z) \, dz = \frac{1}{2 \pi i}\int_\alpha f(w) \, dw \quad .
$$
Since $V$ is simply-connected and $f$ is holomorphic in $V \setminus \{ h(a) \}$, we can apply the Residue theorem to
the right-hand side:
$$
\frac{1}{2 \pi i}\int_\alpha f(w) \, dw = \operatorname{Res}(f, h(a)) I(\alpha, h(a))
$$
and it remains to show that the winding number $I(\alpha, h(a))$ of $\alpha$ with respect to $h(a)$ is one:
$$
 I(\alpha, h(a)) = \frac{1}{2 \pi i}\int_\alpha \frac{dw}{w - h(a)}
 = \frac{1}{2 \pi i}\int_\gamma \frac{h'(z)}{h(z) - h(a)} \, dz
 = \operatorname{Res}(\frac{h'}{h-a}, a)
$$
because $ h'/(h-a)$ is holomorphic in $U \setminus \{ a \}$.
Finally, this residue is one since $h$ is injective, so that
$$
 h(z) = h(a) + h'(a)(z-a) + O((z-a)^2) \\
\implies \frac{h'(z)}{h(z)-h(a)} = \frac{1}{z-a} + O(1)
$$
for $z \to a$.

So you were quite close. We have
$$
\frac{1}{2 \pi i}\int_\alpha f(w) \, dw = \operatorname{Res}(f, h(a)) 
$$
because $\alpha$ circles around $h(a)$ exactly once, and because we
can assume the image to be simply connected.
Also
$$
\frac{1}{2 \pi i}\int_\alpha f(w) \, dw = \frac{1}{2 \pi i}\int_{|w-h(a)| = \epsilon} f(w)
$$
holds for small enough $\epsilon$ because of Cauchy's integral theorem.
