Bundle isomorphisms for $J$-holomorphic tangent bundle

Yes, the domains of $\pi_{1,0}$ and $\pi_{0,1}$ should be $TM$, not $TM\otimes\mathbb{C}$. This is the case for the version on her website; see here, page 78.

Note that $TM$ is a real vector bundle, and it becomes a complex vector bundle by defining $(a + bi)\cdot v := av + bJ(v)$. This is what we mean when we say $(TM, J)$ is a complex vector bundle.

With this in mind, we have $\pi_{1,0} : TM \to T_{1,0}$ and

$$\pi_{1,0}(i\cdot v) = \pi_{1,0}(J(v)) = (\pi_{1,0}\circ J)(v) = i\pi_{1,0}(v) = i\cdot\pi_{1,0}(v).$$

So the real vector bundle isomorphism $\pi_{1,0}$ is also complex linear, and hence is a complex vector bundle isomorphism.

On the other hand, we have $\pi_{0,1} : TM \to T_{0,1}$ and

$$\pi_{0,1}(i\cdot v) = \pi_{0,1}(J(v)) = (\pi_{0,1}\circ J)(v) = -i\pi_{0,1}(v) = i\cdot\pi_{0,1}(v)$$

where the action in the last expression is using the action of $\mathbb{C}$ on $\overline{T_{0,1}}$ (where $i$ acts as multiplication by $-i$). So $\pi_{0,1} : TM \to \overline{T_{0,1}}$ is an isomorphism of complex vector bundles.

Therefore $TM \cong T_{1,0} \cong \overline{T_{0,1}}$.