Question 1
First of all, when you discuss chiral symmetry spontaneous breaking, you need to assume pure QCD theory.
QCD lagrangian with $u-,d-,s-$quarks (they have relatively small masses in compare with $b, t, c$-quarks) has the form
$$
\tag 1 L_{QCD} = \bar{q}_{i}i\gamma_{\mu}D^{\mu}_{ij}q_{j} - \frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - \bar{q}_{ij}M^{ij}q_{j}
$$
In high energy limis, when we may neglect the mass term in compare with kinetic term, $(1)$ is invariant under combined vector-axial $SU_{V}(3)\times SU_{A}(3)$ global transformation, precisely
$$
q \to e^{iq_{V}t_{a}\epsilon_{a} + iq_{A}t_{a}\theta_{a}}q,
$$
where $t_{a}$ is $3\times 3$ Gell-Mann matrices. Here I don't take into account broken by QCD anomaly vector-axial $U(1)$ part of restored symmetry.
By introducing left-right chirality projectors, we may state that
$$
\tag 2 SU_{V}(3)\times SU_{A}(3)\sim SU_{L}(3)\times SU_{R}(3)
$$
To summarize, we have important statement that $SU_{L}(3)\times SU_{R}(3)$ becomes exact symmetry at high energies. $SU_{L}(3)\times SU_{R}(3)$ group is called chiral symmetry group of QCD lagrangian.
This statement implies existense of extra physical states which have same values of spin, strangeness and the baryon number as experimentally observed ones, but with opposite parity. Since we don't see these states, but below energy scale $E\sim \Lambda_{QCD} \sim 0.1 \text{ GeV}$ we see approximately massless meson states, we have to require that aproximate $SU_{L}(3)\times SU_{R}(3)$ symmetry is spontaneously broken in QCD by itself down to $SU_{\text{diag}}(3)$ at scales $\sim \Lambda_{QCD}$. It is broken by quark bilinear form VEV, namely $\langle |\bar{u}u|\rangle \sim \Lambda_{QCD}^3$. In the result, we have pseudogoldstone bosons octet - $\pi^{0, \pm}, \eta^{0} , K^{0}, \bar{K}^{0, \pm}$.
Suppose now we neglect the $s-$quark contribution into such picture. Then we deal with SSB $SU_{L}(2)\times SU_{R}(2)$ down to $SU_{\text{diag}}(2)$. $SU_{\text{diag}}(2)$ in this case is the group of isospin transformation. It is not exact, however since masses of $u-,d-$quarks are not the same: for example, this drives the pure QCD reaction
$$
d + d \to He^{4} + \pi^{0}
$$
which doesn't conserve the isospin.
Questions 2 and 3
QCD chiral symmetry in principle has nothing common to electroweak symmetry. First of all, QCD chiral symmetry group is global symmetry of quark triplets transformation, left and right, while electroweak symmetry group is combined local $SU(2)$ transformation of left quark weak doublets and their local $U_{Y}(1)$ trasformation:
$$
\begin{pmatrix} u \\ d\end{pmatrix} \to e^{i\frac{1-\gamma_{5}}{2}Q_{L}\sigma_{a}\epsilon_{a}(x) + iQ_{Y}\theta (x)}\begin{pmatrix} u \\ d\end{pmatrix}
$$
Electroweak symmetry possesses interactions, while chiral QCD symmetry just state global conservation of chiral charges (i.e., there isn't associated interaction). But if we include electroweak symmetry to $(1)$, then even if we completely neglect the quark masses, the chiral symmetry becomes explicitly (on the level of action), not spontaneously (on the level of states which is described by action) broken because of chiral nature of electroweak group. For example, electromagnetic breaking of isospin symmetry, which breaks $SU(2)$ isospin symmetry down to symmetry transformation generated by $\sigma_{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$, makes the contribution in the masses of $\pi^{\pm}$ mesons, so that they become different from the mass of $\pi^{0}$ meson, while QCD predicts precisely exact equality of these masses.
Next, electroweak group symmetry is also spontaneously broken at scale $\approx 240\text{ GeV}$, namely
$$
SU_{L}(2)\times U_{Y}(1) \to U_{EM}(1),
$$
by VEV of Higgs field doublet $\langle|H |\rangle \sim v$. The scales $\Lambda_{QCD}, v$ are different, and in general has different nature.
Finally, the fact that only left charged currents (i.e., $\bar{u}\gamma_{\mu}\left(\frac{1- \gamma_{5}}{2}\right)d$) interact with electroweak sector comes from the experiment. Your statement that only left fermions interact electroweakly is incorrect, since all electrically charged right particles interact with EM field as well as left, and all right particles interact with $Z-$boson too.
Best Answer
It's actually the other way around. That axial rotation of the pions ensures that, given its non-vanishing v.e.v., given by the condensate (assumed to be produced by QCD: a fact!), they therefore must be the Goldstone modes of the SSB of the axial charges. But, first, the chiral condensate is required so as to unleash all this.
Take L=2, for simplicity, and let's be schematic (~) about normalizations, which you may adjust to your satisfaction, in comportance with the conventions of your text.
Let us consider the relevant fermion bilinears and their representation of the $SU(2)_L\times SU(2)_R$. (By the way, the 3 axial $\vec{Q}_A$ do not close to an SU(2), as you wrote, as their commutators close to $SU(2)_V$ instead. Don't ever write this again... Also, $U(1)_A$ is broken explicitly by the anomaly, not spontaneously.) So, the 4 bilinears, $\bar{\psi} {\psi} $, $\bar{\psi}\gamma_{5} \vec{\tau} {\psi} $ form a quartet of this chiral group formally analogous of the $\sigma, \vec{\pi}$ quartet of the σ-model of the 60s; in fact, they are the QCD interpolating fields for this quartet.
(If you are insouciant about i s and the like, you may think of this scalar-pseudoscalars quartet as a 4-vector $t,\vec{x}$ of the Lorentz group, as a familiar mnemonic of the combinatorics/groupery to follow: The vector isorotations are analogous to the 3 rotations, and the 3 axials are analogous to the 3 boosts, acting on 4-vectors.)
You then see that $$ \left[\vec{Q}_{V},\bar{\psi}\psi\right]=0, $$ $$ \left[\vec{Q}_{A},\bar{\psi}\psi\right]\sim \bar{\psi} \gamma_{5} \vec{\tau} \psi ~, $$ $$ \left[Q_{a}^{V} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim \epsilon _{abc} \bar{\psi}\gamma_{5}\tau^c\psi ~, $$ and, crucially, the relation of interest, where you note the $\gamma_{0}\gamma_{5}$ makes all the difference in the combinatorics, as in the σ -model link, above, $$ \left[Q_{a}^{A} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim \delta_{ab} \bar{\psi}\psi ~. $$
So the σ is an isosinglet; the axials transform the σ by $\vec\pi$s; the vector transform of the $\vec \pi$s is an isorotation thereof; and the suitable "diagonal" axial transforms of the $\vec\pi$s takes them to the σ, the QCD analog of the Higgs of the EW interactions, the guy with the v.e.v.
Now take the v.e.v. $ \langle \Omega \lvert ... ... \rvert\Omega \rangle$ of each of the above. The r.h.sides of the first 3 must vanish; the v.e.v.s of the Goldstone modes are null!
The v.e.v. of the last one does not, but $=v\approx (250MeV)^3$, your relation of interest. QCD just achieves that, by dint of strong dynamics. So, over and above making it impossible for the axials to annihilate the vacuum, it identifies the 3 pions as the Goldstone modes of the 3 SSBroken $\vec{Q}_A$.
In point of fact, you actually see that $\vec{Q}_{A} \rvert\Omega \rangle \sim |\vec{\pi}\rangle$, that is the axial charges pump pions (chiral goldstons) out of the vacuum--- the precursor to PCAC.
A final loose end, lest you might still object that the third relation with vanishing r.h.side would then be moot, $\langle \vec{\pi} \lvert \bar{\psi} \psi \rvert\Omega \rangle=0$; but, no, the pion is orthogonal to the σ, just as it is to the original vacuum "chosen", $\langle \vec{\pi} \vert\Omega \rangle=0$. (Could insert $\vert\Omega\rangle \langle\Omega\vert$ above and factor out the order parameter, $\langle \vec{\pi} \vert\Omega \rangle v=0$ )
Further related (formally identical) questions might be 281696, and, needless to say, Gell-Mann and Lévy's timeless 1960 σ -model classic.