Suppose that $M$ is an uncountable and saturated model of a given totally transcendental theory and let $A$ be a countable subset of $M$. Assume that $M_1 \subset M$ and $M_2 \subset M$ are two arbitrary prime extensions of the set $A$, in particular countable. I know that every countable subset of M has a prime extension since the theory is totally transcendental, therefore there exists a model $N \subset M$ which is a prime extension of $M_1 \cup M_2$. Now I want to argue that $N$ is a prime extension of $A$. But I think that this follows from the fact(?) that constructible extensions and prime extensions are the same thing and that constructabiliy is transitive: $N$ is prime, hence constructible, over $M_1 \cup M_2$ and $M_1 \cup M_2 $ is constructable over $A$ because $M_1$ and $M_2$ are prime, hence constructable, over $A$.
Is this reasoning correct? And somehow I wonder why there is this notion of constructability if its the same as being prime?
Your reasoning is not quite correct. The gap is where you claim that $M_1\cup M_2$ is constructible over $A$ because $M_1$ and $M_2$ are constructible over $A$. Recall that $B$ is constructible over $A$ if there is an enumeration of $B$, $(b_\alpha)_{\alpha<\gamma}$ (indexed by an ordinal $\gamma$) such that for all $\alpha<\gamma$, $\mathrm{tp}(b_\alpha/A(b_\beta)_{\beta<\alpha})$ is isolated. If $M_1$ and $M_2$ are constructible over $A$, we have construction sequences $(m_\alpha)_{\alpha<\gamma}$ and $(m'_\alpha)_{\alpha<\gamma'}$ for $M_1$ and $M_2$, respectively. But you can't, in general, put these together to form a construction sequence for $M_1\cup M_2$. For example, there might be $m\in M_1$ and $m'\in M_2$ such that $\mathrm{tp}(mm'/A)$ is not isolated. Then every model containing $M_1\cup M_2$ fails to be atomic over $A$, and hence fails to be constructible over $A$.
Now let me address your more general question: "I wonder why there is this notion of constructibility if it's the same as being prime?"
One reason is that the definition of "prime" is less technical / more natural, while the definition of "constructible" is often useful for carrying out more fine-grained proofs by transfinite induction.
Another reason is that the notions of prime and constructible don't agree in all contexts. In any situation, constructible over $A$ implies prime over $A$. The converse is true (so constructible over $A$ is equivalent to prime over $A$) when (a) the set $A$ and the language $L$ are both countable, or (b) $T$ is totally transcendental. But I'll remark that that the implication from prime to constructible when $T$ is totally transcendental is quite a bit trickier to prove than the rest of the facts here. If $T$ is not totally transcendental and $A$ or $L$ is uncountable, there can be models which are prime over $A$ but not constructible over $A$. For more on these notions in arbitrary theories, and the relationship with "atomic", see my answer here.
One last comment: you wrote "I know that every countable subset of $M$ has a prime extension since the theory is totally transcendental". In fact, in a totally transcendental theory, every set has a constructible (and hence prime) extension, with no cardinality restriction.