First note that if $\omega$ is a $(p, q)$-form, $J^*\omega = i^{p-q}\omega$. From now on, if $\omega \in \mathcal{E}^r(X)\otimes_{\mathbb{R}}\mathbb{C}$, I will write $\omega^{p,q}$ for the $(p, q)$ part of $\omega$.
Suppose $r = 1$. If $\omega \in \mathcal{E}^1(X)\otimes_{\mathbb{R}}\mathbb{C}$,
\begin{equation}
\omega = \omega^{1, 0} + \omega^{0,1}.\tag{1.1}
\end{equation}
Note that $J^*\omega = J^*\omega^{1,0} + J^*\omega^{0,1} = i\omega^{1,0} - i\omega^{0,1}$, so
\begin{equation}
iJ^*\omega = -\omega^{1,0} + \omega^{0,1}.\tag{1.2}
\end{equation}
Using equations $(1.1)$ and $(1.2)$, we obtain
\begin{align*}
\omega^{1,0} &= \frac{1}{2}(\omega-iJ^*\omega)\\
\omega^{0,1} &= \frac{1}{2}(\omega+iJ^*\omega).
\end{align*}
Now consider the case $r = 2$. If $\omega \in \mathcal{E}^2(X)\otimes_{\mathbb{R}}\mathbb{C}$,
\begin{equation}
\omega = \omega^{2,0} + \omega^{1,1} + \omega^{0,2}. \tag{2.1}
\end{equation}
Note that
\begin{equation}
J^*\omega = J^*\omega^{2,0} + J^*\omega^{1,1} + J^*\omega^{0,2} = -\omega^{2,0} + \omega^{1,1} -\omega^{2,0}.\tag{2.2}
\end{equation}
Using equations $(2.1)$ and $(2.2)$, we obtain
\begin{align*}
\omega^{2,0} + \omega^{0,2} &= \frac{1}{2}(\omega - J^*\omega)\\
\omega^{1,1} &= \frac{1}{2}(\omega + J^*\omega).
\end{align*}
In this case, we can't use $J^*$ to distinguish between $(2, 0)$-forms and $(0,2)$-forms so we can't isolate them in a formula. The issue is that both $(2, 0)$-forms and $(0, 2)$-forms are eigenvectors of $J^*$ with eigenvalue $-1$. As the only complex two-forms which are eigenvectors of $J^*$ with eigenvalue $1$ are $(1, 1)$-forms, we get the formula for $\omega^{1,1}$.
Now let's consider the general case.
If $r = p + q$ is odd, then $p - q$ is odd, so $i^{p-q} = \pm i$. In this case, $iJ^*$ has eigenvalues $\pm 1$, and
$$iJ^*\omega^{p,q} = \begin{cases}
-1 & p - q \equiv 1 \bmod 4\\
1 & p - q \equiv 3 \bmod 4.
\end{cases}$$
Therefore
\begin{align*}
\sum_{\substack{p+q = r\\ p - q\ \equiv\ 1 \bmod 4}}\omega^{p,q} &= \frac{1}{2}(\omega - iJ^*\omega)\\
\sum_{\substack{p+q = r\\ p - q\ \equiv\ 3 \bmod 4}}\omega^{p,q} &= \frac{1}{2}(\omega + iJ^*\omega).
\end{align*}
When $r = 1$, there is only one possible pair $(p, q)$ with $p - q \equiv 1 \bmod 4$ (namely $(p, q) = (1, 0)$), and only one possible pair with $p - q \equiv 3 \bmod 4$ (namely $(p, q) = (0, 1)$). This is why we get the explicit formulae for these forms. For all other odd $r$, there is more the one pair $(p, q)$ with $p - q \equiv 1 \bmod 4$ and more than one pair with $p -q \equiv 3 \bmod 4$, so we don't get an explicit formula for any of the $(p, q)$-parts.
If $r = p + q$ is even, $i^{p - q} = \pm 1$. In this case, $J^*$ has eigenvalues $\pm 1$ and
$$J^*\omega^{p,q} = \begin{cases}
-1 & p - q \equiv 2 \bmod 4\\
1 & p - q \equiv 0 \bmod 4.
\end{cases}$$
Therefore
\begin{align*}
\sum_{\substack{p+q = r\\ p - q\ \equiv\ 2 \bmod 4}}\omega^{p,q} &= \frac{1}{2}(\omega - J^*\omega)\\
\sum_{\substack{p+q = r\\ p - q\ \equiv\ 0 \bmod 4}}\omega^{p,q} &= \frac{1}{2}(\omega + J^*\omega).
\end{align*}
When $r = 2$, there is more than one possible pair $(p, q)$ with $p - q \equiv 2 \bmod 4$, but there is only one possible pair with $p - q \equiv 0 \bmod 4$ (namely $(p, q) = (1, 1)$). This is why we get the explicit formula for $\omega^{1,1}$ but not for $\omega^{2, 0}$ or $\omega^{0,2}$. For all other even $r$, there is more the one pair $(p, q)$ with $p - q \equiv 2 \bmod 4$ and more than one pair with $p -q \equiv 0 \bmod 4$, so we don't get an explicit formula for any of the $(p, q)$-parts.
In summary, we have
\begin{align*}
\omega^{1,0} &= \frac{1}{2}(\omega - iJ^*\omega)\\
\omega^{0,1} &= \frac{1}{2}(\omega + iJ^*\omega)\\
\omega^{1,1} &= \frac{1}{2}(\omega + J^*\omega),\\
\end{align*}
but we have no such formulae for any other $\omega^{p,q}$.
You may be interested in my related question (and its answer): When is a $k$-form a $(p, q)$-form?
