How can I tell that a Lagrangian has an $SU(2)\times SU(2)$ symmetry?

You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc... and stick to the invariance of the bilinear of scalars, at the heart of your puzzlement.

I suspect you are asking how is the bilinear $\phi^\dagger \cdot \phi$ invariant under two different, commuting SU(2)s instead of the familiar gauged SU(2) left-acting on the complex vector, \begin{equation} \phi \equiv\begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix} . \end{equation} First you need to recall that SU(2) is pseudoreal, i.e. the conjugate representation $\tilde{\phi}\equiv i\tau_2 \phi^*$ is equivalent to this fundamental one, that is, acting on \begin{equation} \tilde{\phi} = \begin{pmatrix} \phi^*_{2}\\- \phi^*_1 \end{pmatrix} \end{equation} produces the same transformation on all four components of $\phi$ , real and imaginary pieces.

Now consider the 2x2 complex matrix with columns $\tilde{\phi}$ and $\phi$, respectively, so $\Phi\equiv \frac{1}{\sqrt{2}} [ \tilde{\phi} ~,~ \phi ]$, so $$ \Phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi^*_{2}& \phi_{1} \\ -\phi^*_{1}& \phi_{2} \end{pmatrix} ~, $$ and check that $$\Phi^\dagger \Phi = (\phi^\dagger \!\cdot \phi)~ 1\!\!\!1 ~/~2 ~,$$ so $\operatorname {Tr} \Phi^\dagger \Phi = \phi^\dagger\! \cdot \phi$.

Now, left multiplication of $\Phi$ by a unitary SU(2) rotation leaves the matrix bilinear of $\Phi$s invariant, and, a fortiori, its trace invariant.

Significantly, however, right multiplication by another, independent SU(2) which knows nothing about the left one, scrambles the two columns of $\Phi$ with each other, preserving, however their left-SU(2) transformation properties, since, as we saw, each column of $\Phi$ amounts to the same outcome when left-SU(2) transformed. It is evident, but you may convince yourself by a right transformation, a left transformation, and then the inverse right transformation--you will be left with the original left transformation.

Even though the $\Phi$ bilinear is not right-SU(2) invariant, by the cyclicity of the trace, its trace is. So all $\phi^\dagger \cdot \phi$ bilinears are both left, but also right SU(2) invariant, and so are all lagrangian kinetic and potential terms you'd construct.The right-invariance, however, will be spoiled by coupling to gauge fields, as you might check.