Disclaimer: I am not as familiar with differential topology as I am with algebraic topology.
I believe your question can be answered via the classification of covering spaces. I think it can also be done by invoking various lifting/uniqueness criteria regarding covers, but this way is less technical.
(I am being sloppy and ignoring basepoints/additional conditions)
Two covers $p_1 : X_1 \to X$, $p_2:X_2 \to X$ are said to be isomorphic if there is some homeomorphism $f:X_1 \to X_2$ such that $p_2 \circ f = p_1$. This agrees with the intuitive notion of two covers being "essentially the same".
The classification of covering spaces states (among other things) that there is a bijection
$$\{\text{isomorphism classes of path connected covers of }X \} \leftrightarrow \{\text{conjugacy classes of subgroups of } \pi_1(X,x_0) \}$$
Where an isomorphism class of a covering space $p:X_1 \to X$ is associated with the image of its fundamental group under the induced map $p_\ast$.
Furthermore, if $H \subseteq G \subseteq \pi_1(X,x_0)$, with corresponding covers $p_H:X_H \to X$, $p_G:X_G\to X$, then there is some cover $f:X_H \to X_G$ with $p_G \circ f = p_H$.
Let $p:E \to B$ be some cover. If $E$ is simply connected, then the image of its fundamental group is trivial (note that this makes 80.3 a special case of the classification). The trivial subgroup is of course its own conjugacy class, so it is associated to a single isomorphism class of covers. From this we can conclude that any two covers with simply connected domains must be isomorphic, and in particular your $q,q'$ are homeomorphisms.
Does this solve your problem?
First the isomorphism between the complex bundle $(T_{\Bbb{R}}X,I)$ and $T^{(1,0)}X$ is really the linear algebra, pick the frame $$(T_{\Bbb{R}}X,I) \to T^{(1,0)}X\\\frac{\partial}{\partial x_i}\mapsto \frac{1}{2}(\frac{\partial}{\partial x_i } - i \frac{\partial}{\partial y_i})$$
If can be checked this is a complex isomorphism, as the local frame is isomorphic the vector bundle is also isomorphic.
For the first one and the third one :
Since $$\frac{\partial}{\partial z_i} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} -i\frac{\partial}{\partial y_i}\right)$$
We know in the real case $$\frac{\partial}{\partial x_j} = \frac{\partial \tilde{x}_j}{\partial x_i} \frac{\partial}{\partial \tilde{x_j}}$$
Similar for $y_i$,
Substitute into the above definition gives that $$\frac{\partial }{\partial z_i} = \frac{\partial \tilde{z}_j}{\partial z_i} \frac{\partial}{\partial \tilde{z_j}}$$
Therefore the cocycle is really given in the first definition,
therefore we see these three definitions are equivalent.
Maybe these three different characterizations have their own strength
for example the cocycle characterization will be useful in building some abstract theory about vector bundles (for example when dealing with classification problems).
The third one is very useful in local computation since we are familiar with differential calculus on the local coordinate for the real case, the calculation will naturally extend to the complexified case in the third setting, that's why this is useful.
Best Answer
Triviality of $TM$ does not imply that of $TN$. For example, let $M=\mathbb R^2$, and let $\mathbb Z$ act on $M$ by $$ n\cdot (x,y) = (x+n,(-1)^n y). $$ Let $N$ be the quotient space $M/\mathbb Z$, and let $p:M\to N$ be the quotient map. Then $p$ is a smooth covering map, and $N$ is diffeomorphic to the open Möbius band, which is not even orientable, let alone parallelizable.