Poisson summation formula and periodic summation of Fourier transforms
But we can choose the functions $x_1(t)$ and $x_2(t)$, even though they are equal at the sample points, in such a way that their transforms $X_1(f)$ and $X_2(f)$ are quite different from each other, and, therefore, so will their periodizations. 


That's a wrong conclusion. If $x_1(nT)=x_2(nT)$ $\forall n$ then the discrete-time Fourier transforms of the respective sequences must be identical, and so must be their periodized spectra.

Note that $x_1(nT)=x_2(nT)$ and $x_1(t)\neq x_2(t)$ implies that at least one of the two functions is not sampled according to the sampling theorem, i.e., the function is not completely represented by its samples. Because otherwise equal samples would imply equal functions. This might be the source of your confusion. If both functions are band-limited and if the sampling theorem is satisfied then $x_1(nT)=x_2(nT)$ implies $x_1(t)=x_2(t)$.