Transformation of field operator under SU(2) x SU(2) I am a little confused on how field operators transform under compound symmetry groups. The following text is copied from Michael Dine Supersymmetry and String Theory

As an example, relevant both to the strong and the weak interactions,
  consider a theory with a symmetry $SU(2)_L \times SU(2)_R$. Take $M$ to
  be a Hermitian, matrix field, 
  $$M = \sigma + i\vec{\pi} \cdot\vec{\sigma}.$$ 
  Under the symmetry, which we first take to be global,
  $M$ transforms as $$ M \rightarrow g_LMg_R $$ with $g_L$ and $g_R$ are SU(2) matrices.

How does the transformation of $M$ follow? If the symmetry group were one single SU(2), I would understand that $M$ transforms as
$$ M \rightarrow g^\dagger M g. $$
Generalizing to SU(2)$_L$ × SU(2)$_R$, I would guess that $M$ transforms as
$$ M \rightarrow g_R^\dagger g_L^\dagger M g_L g_R. $$
I don't think my naive guess is consistent with the formula in Dine's book.
My further question is, supposing the field operator transforms in the fundamental representation, how does it transform under $SU(2) \times SU(2)$?
 A: Mike is displaying a chiral model, the best known σ-model, which is the prototype for global symmetry realizations and spontaneous symmetry breaking. 
The (unnormalized) unitary matrix M he defines through the spinor map actually transforms like the fundamentals, not the adjoint, of both SU(2)s, L and R.
To see this, note left multiplication by gL transforms each column of M as a complex doublet; while right-multiplication by gR transforms each row of it as a complex doublet. So far, the two SU(2)s commute, and have no clue about each other. $g_L\otimes g_R$ means the Ls and the Rs act on different vector spaces and so cannot multiply each other.
However, suitable linear combinations of them, e.g. the vector combination (as per the WP reference provided) amount to the isospin "vector" SU(2) subgroup of them. I.e. if we take the rotation angle (parameter) of both the left and the right groups to be the same, (think of synchronized swimming), you then have your adjoint action of that isospin group, as written,
$$ M \rightarrow g^\dagger M g. $$
The bifundamental rep of the full group is, ipso facto, the adjoint of the diagonal subgroup.
You may see that the σ remains invariant, and the three π s transform in the adjoint of the isospin V .
(Never again write the following, meaningless,  equation you wrote: Lefters act from the left, and Righters from the right. Again: they cannot multiply each other.)
If you took the parameters of left and right to be opposite each other, the action you'd get would actually be nonlinear, but the 3 Axial generators A of SO(4)~SU(2)×SU(2) this represents do not close into a subalgebra of SO(4).
Each connects the σ  with the triplet of π s. (They represent a scalar meson and the lightest pseudo scalar mesons, respectively, of course). 
By the way, you might have chosen to write the σ and the three π s of your complex doublet as a real 4-vector, the fundamental of your full group, SO(4),
$$ \begin{pmatrix} {\boldsymbol{ \pi}} \\ \sigma \end{pmatrix} \stackrel{SO(4)}{\longrightarrow} \begin{pmatrix} {\boldsymbol{ \pi}'} \\ \sigma' \end{pmatrix} = \left[ \mathbf{1}_4+ \sum_{i=1}^3 \theta_i^V V_i + \sum_{i=1}^3 \theta_i^A A_i \right] \begin{pmatrix} {\boldsymbol{ \pi}} \\ \sigma \end{pmatrix} $$
where 
$$ \sum_{i=1}^3 \theta_i^V V_i =\begin{pmatrix}
0 & -\theta^V_3 & \theta^V_2 & 0 \\
\theta^V_3 & 0 & -\theta_1^V & 0 \\
-\theta^V_2 & \theta_1^V & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\qquad \qquad 
\sum_{i=1}^3 \theta_i^A A_i = \begin{pmatrix}
0 & 0 & 0 & \theta^A_1 \\
0 & 0 & 0 & \theta^A_2 \\
0 & 0 & 0 & \theta^A_3 \\
-\theta^A_1 & -\theta_2^A & -\theta_3^A & 0 
\end{pmatrix}.$$
There is no substitute to doing all 3 & 3 transformations of the SO(4) and monitoring their effect on every field. A related question displaying such (U ~ M here) is 304811, or 102690/66086.

Frivolous note added on the toxic original notation: I corrected the formula in the question to reflect the original (2.39) of Mike's text. However, as it stands, here and in the book, it is still breathtakingly inviting confusion. 
The point is, from time immemorial, the community uses $\vec{~\tau}$ for the Pauli matrix triplet ($\vec{\sigma}$) when denoting isospin instead of spin, precisely to avoid confusions with the scalar field σ multiplying the identity. So, in most texts, the formula reads: $M=\sigma+ i\vec{\pi}\cdot\vec{\tau}$, the most general group element of SU(2) in the doublet representation, up to an overall normalization;  σ being a scalar   field and π being a 3-vector field. I only caught on to the abusive notation by reading the commentary, too late. This is the classic bad choice/typo invisible to people in the know that is hard to notice, but which stresses new readers...
