Summary: Neither of your two approaches is in accordance with a correct application of a $\pi/2$-pulse. The first matrix contains a fabricated phase shift, the second one is obtained by a basis change which can't simply be applied twice.
Some introductory remarks
- A $\pi/2$-pulse represents a 90°-rotation of the Bloch vector on the Bloch sphere around some axis. Around which axis depends on the form of the interaction Hamiltonian, which should be of the form $V \simeq \Omega_x\sigma_x + \Omega_y \sigma_y $, from which you derive the unitary transformation above by changing into a suitable interaction picture etc.. It is this unitary that should be used.
- Furthermore, having a $\pi/2$-pulse puts certain restrictions on the length of the pulse, i.e. you get a certain condition on the length of the pulse, which is $\Omega t = \pi$ if I'm not mistaken. So when you write $U_{"\pi/2"}$, you should have eliminated already all $\Omega$'s and $t$'s there, leaving you with something like
$$ U_{"\pi/2"} = \frac{1}{\sqrt 2} \begin{pmatrix}1&1\\-\mathrm i&\mathrm i\end{pmatrix},$$
and some expression accordingly for $U_{"\pi/2"}'$. (I will later show why $U_{"\pi/2"}$ is not a $\pi/2$ pulse, therefore the $"\pi/2"$ in it.)
Non-equivalence of the two matrices
What you do here is to decompose the unitary into $U'_{"\pi/2"}$ and a second unitary matrix of the form
$$ U_b = \begin{pmatrix}1&0\\0&\mathrm i\end{pmatrix},$$
which also corresponds to a basis change of your state vector. This is equivalent to introduce a phase between the two basis states, or to rotate the basis of the Bloch sphere accordingly. However, if you apply two pulses $U_{"\pi/2"}$ in row,
$$U_{"\pi/2"} U_{"\pi/2"} = U_{"\pi/2"}' U_b U_{"\pi/2"}' U_b$$
you see that this is inequivalent to two consequential applications of $U_{"\pi/2"}'$ obviously.
Why your $U_{"\pi/2"}$ is not a $\pi/2$-pulse
Coming back to the question why $U_{"\pi/2"}$ applied twice does not give a $\pi$-rotation. This is easy to see now by just multiplying it with itself, yielding a matrix with all entries equal in modulus $(\simeq \pm 1 \pm \mathrm i)$. However, multiplying $U_{"\pi/2"}'$ with itself gives the desired $\pi$-pulse, which requires to have only entries in the off-diagonal.
So what, then, is $U_{"\pi/2"}$ really doing? It is bringing your state vector into a superposition state when starting from one of the two basis states. But besides a rotation round an axis in the equatorial plane it introduces a phase, which is equivalent to a rotation around the $z$-axis. Thus, your $U_{"\pi/2"}$ is a combination of a $\pi/2$-pulse around, say, the $x$-axis with a rotation around the $z$-axis. A second rotation around the $x$-axis then has a different effect (here, if I see it correct, it does nothing, because the vector is just parallel to the rotation axis), and the further phase (=rotation around the $z$-axis) leads to a final state, which is still in the equatorial plane of the Bloch sphere.
Physically, this situation is similar to that of the frequency of the radiation field being not resonant with the transition frequency of the two-level system. Then, the transformation in the interaction picture does not remove the term $\sim \sigma_z$ of the two-level system Hamiltonian, so that these terms show up in the unitary, yielding a phase shift which is proportional to $\sim \Delta t$, where $\Delta$ is the detuning between the radiation frequency and the transition frequency. However, as the other answer pointed out, the limit of $t=0$ yields some instantaneous phase shift.
As I pointed out in the comment below, this is not unphysical per se, but can accommodate a phase shift between two pulses that is, for instance, imposed by shift of the phase of the radiation field,
which can't be described by a Hamiltonian anyway. A inclusion of such a phase shift may seem unphysical for not yielding the identity transformation for $t \rightarrow 0$, but it can't be described by a Schrödinger equation anyway but should be considered as a kind of external condition.
Let's temporarily forget that the two $m=0$ states exist, and consider just the two completely aligned triplet states,
$\newcommand\ket[1]{\left|{#1}\right>}
\newcommand\up\uparrow
\newcommand\dn\downarrow
\newcommand\lf\leftarrow
\newcommand\rt\rightarrow
$
$\ket{\up\up}$ and $\ket{\dn\dn}$.
There's not any physical difference between these: you can "transform" your state from one to the other by changing your coordinate system, or by standing on your head. So any physical observable between them must also be the same.
Either of the single-particle states are eigenstates of the spin operator on the $z$-axis,
$$\sigma_z = \frac\hbar 2\left(\begin{array}{cc}1&\\&-1\end{array}\right),$$
and "standing on your head," or reversing the $z$-axis, is just the same as disagreeing about the sign of this operator.
But let's suppose that, on your way to reversing the $z$-axis, you get interrupted midway. Now I have a system which I think has two spins along the $z$-axis, but you are lying on your side and think that my spins are aligned along the $x$-axis. The $x$-axis spin operator is usually
$$\sigma_x = \frac\hbar 2\left(\begin{array}{cc}&1\\1\end{array}\right).$$
Where I see my single-particle spins are the eigenstates of $\sigma_z$,
$$\ket\up = {1\choose0} \quad\text{and}\quad \ket\dn = {0\choose1},$$ you see those single-particle states as eigenstates of $\sigma_x$,
\begin{align}
\ket\rt &= \frac1{\sqrt2}{1\choose1}
= \frac{\ket\up + \ket\dn}{\sqrt2}
\\
\ket\lf &= \frac1{\sqrt2}{1\choose-1}
\end{align}
If you are a $z$-axis chauvinist and insist on analyzing my carefully prepared $\ket{\rt\rt}$ state in your $\up\dn$ basis, you'll find this mess:
\begin{align}
\ket{\rt\rt} = \ket\rt \otimes \ket\rt
&= \frac{\ket\up + \ket\dn}{\sqrt2} \otimes \frac{\ket\up + \ket\dn}{\sqrt2}
\\
&= \frac{\ket{\up\up}}2 + \frac{\ket{\up\dn} + \ket{\dn\up}}2 + \frac{\ket{\dn\dn}}2
\end{align}
This state, which has a clearly defined $m=1$ in my coordinate system, does not have a well-defined $m$ in your coordinate system: by turning your head and disagreeing about which way is up, you've introduced both $\ket{\up\up}$ and $\ket{\dn\dn}$ into your model. You've also introduced the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$.
And this is where the symmetry argument comes in. The triplet and singlet states are distinguishable because they have different energies.
If you propose that the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$ is the singlet state, then you and I will predict different energies for the system based only on how we have chosen to tilt our heads. Any model that says the energy of a system should depend on how I tilt my head when I look at it is wrong. So the $m=0$ projection of the triplet state must be symmetric, in order to have the same symmetry under exchange as the $m=\pm1$ projections.
Best Answer
Are you sure that's what the book is asking you to find?
$\hbar/2$ is the eigenvalue of the $S_{x}$ operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the $\mid S_{x};+\rangle$ ket in the $S_{z}$ basis, then you're nearly correct, just a minor sign error:
$$\mid S_{x};+\rangle = \frac{1}{\sqrt{2}}\mid+\rangle + \frac{1}{\sqrt{2}}\mid-\rangle$$