TQBF PSPACE-COMPLETE : Why this algorithm is exponential but Savitch’s not?

So this is a question pertaining to the proof for $ PSPACE-COMPLETE$ (for TQBF for example). The idea is to first prove the $ L$ $ is$ $ PSPACE$ (easy part) and next is to prove $ PSPACE-COMPLETE$ . The latter requires demonstrating an algorithm which computes the L in polynomial space. This is usually achieved by having recursive calls such that is re-used.

In TBQF proof, the equation $ \phi_{i+1}(A,B)$ = $ \exists Z [\phi_{i+1}(A,Z) \land \phi_{i+1}(Z,B) ]$ ($ Z$ is mid-point )is default recursive relation for computing TBQF truth. In any standard proof, it is said that $ \phi_{i+1}(A,B)$ is computed two times and for $ m$ nodes, this formula explodes hence, other recursive-relation should be used to bound.

However in Savitch’s proof, the recursive relation is $ Path(a,b,t)$ = $ Path(a,mid,t-1)$ AND $ Path(mid,b,t-1)$ accepts then ACCEPT. In proof, it is stated that this relation reuses-spaces.

My Question is Why in TBQF relation space explodes while in Path, it is reused? Both of these relations looks more or less same to me because both refers to i-1 instances and will need space to store them?.