Yes, you made a mistake. The chain rules for the Wirtinger derivatives are, for real-differentiable $g$ and $h$,
\begin{gather}
\frac{\partial (g\circ h)}{\partial z} = \biggl(\frac{\partial g}{\partial w}\circ h\biggr)\cdot \frac{\partial h}{\partial z} + \biggl(\frac{\partial g}{\partial \overline{w}}\circ h\biggr)\cdot \frac{\partial \overline{h}}{\partial z},\\
\frac{\partial (g\circ h)}{\partial \overline{z}} = \biggl(\frac{\partial g}{\partial w}\circ h\biggr)\cdot \frac{\partial g}{\partial \overline{z}} + \biggl(\frac{\partial g}{\partial \overline{w}}\circ h\biggr)\cdot \frac{\partial \overline{h}}{\partial \overline{z}}.
\end{gather}
Also, we have $\frac{\partial \overline{h}}{\partial \overline{z}} = \overline{\frac{\partial h}{\partial z}}$.
So if $f$ is holomorphic, and $u$ twice real-differentiable, from the chain rules we obtain
$$\frac{\partial (u\circ f)}{\partial \overline{z}} = \biggl(\frac{\partial u}{\partial w}\circ f\biggr)\cdot \underbrace{\frac{\partial f}{\partial \overline{z}}}_{= 0} + \biggl(\frac{\partial u}{\partial \overline{w}} \circ f\biggr)\cdot \frac{\partial \overline{f}}{\partial \overline{z}} = \biggl(\frac{\partial u}{\partial \overline{w}} \circ f\biggr)\cdot \overline{f'},$$
and then from the chain and product rules
\begin{align}
\frac{\partial^2 (u\circ f)}{\partial z \partial \overline{z}}
&= \frac{\partial}{\partial z}\Biggl[\biggl(\frac{\partial u}{\partial \overline{w}} \circ f\biggr)\cdot \overline{f'}\Biggr]\\
&= \Biggl[\biggl(\frac{\partial^2 u}{\partial w\partial \overline{w}} \circ f\biggr)\cdot f'\biggr]\cdot \overline{f'} + \biggl(\frac{\partial u}{\partial \overline{w}} \circ f\biggr)\cdot \frac{\partial\overline{f'}}{\partial z}\\
&= \biggl(\frac{\partial^2 u}{\partial w\partial \overline{w}} \circ f\biggr)\cdot \lvert f'\rvert^2,
\end{align}
since $\overline{f'}$ is antiholomorphic. In your computation of $\Delta (u\circ f)$, the first term in the brackets should have been $(u_{z\overline{z}} \circ f)\overline{f}_{\overline{z}} f_z$ instead of $(u_{z\overline{z}}\circ f) f_\overline{z} f_z$; the harmonicity of $u$ is [except for locally constant $f$] essential.
Multiplying by $4$, that becomes
$$\Delta (u\circ f) = \bigl((\Delta u)\circ f\bigr) \cdot \lvert f'\rvert^2.\tag{1}$$
So composition with a holomorphic function in that order preserves harmonicity.
An example shows that composition in the other order generally doesn't preserve harmonicity: consider $f(z) = z^2$ and $u(w) = \operatorname{Re} w$. You made a similar mistake in that computation as in the first, you took only one of the two summands arising from the chain rule for the Wirtinger derivative into account.
The computation yields
\begin{align}
\frac{\partial (f\circ u)}{\partial z \partial \overline{z}}
&= \frac{\partial}{\partial z}\biggl[\bigl(f'\circ u\bigr)\cdot \frac{\partial u}{\partial \overline{z}}\biggr]\\
&= \biggl[\bigl(f''\circ u\bigr)\cdot \frac{\partial u}{\partial z}\biggr]\cdot \frac{\partial u}{\partial \overline{z}} + \bigl(f'\circ u\bigr)\cdot \frac{\partial^2 u}{\partial z\partial \overline{z}},
\end{align}
and hence
$$\Delta (f\circ u) = \bigl(f'' \circ u\bigr)\cdot \frac{\partial u}{\partial z}\cdot \frac{\partial u}{\partial \overline{z}} + \bigl(f'\circ u\bigr)\cdot \Delta u.\tag{2}$$
This shows that for a harmonic $u$ (defined on a connected open subset of $\mathbb{C}$) and a holomorphic $f$ (also defined on a connected open subset of $\mathbb{C}$), the composition $f\circ u$ is harmonic if and only if $\frac{\partial u}{\partial z} \equiv 0$, $\frac{\partial u}{\partial \overline{z}} \equiv 0$, or $f'' \equiv 0$, that is, $u$ must be holomorphic or antiholomorphic, of $f$ must be [the restriction of] a polynomial of degree $\leqslant 1$.
