the matrix inequlity Is it true that $$\|A\| \leq \|A^2\|$$
for $A \in SL(2,\mathbb{R})$, where $\| \|$ is the operator norm that is the first singular value?
$$\left \| A \right \| =\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A).$$
Definitively, it is not true for $GL(2, \mathbb{R})$ as one can consider $A=diag(1/3, 1/3).$

If by "singular values" you mean the diagonal entries in writing $A=k_1 \pmatrix{t&0 \cr 0&t^{-1}}k_2$, with $k_1, k_2$ in the orthogonal group $SO(2)$, then your speculation is correct. EDIT: if you truly mean "spectral norm", then this is not correct. But/and in many contexts "singular values" does not refer to eigenvalues. Dunno... Let me continue under the assumption that you do not actually mean eigenvalues or spectral norm, but as I wrote... (someone may be interested in that scenario, in any case).
EDIT: this asserts nothing about diagonalizability of $A$. The general version is "Cartan decomposition". Then, letting $\delta=\pmatrix{a&0\cr 0&d}$,
$$
A^\top A \;=\; (k_1 \delta k_2)^\top(k_1Ak_2)
\;=\; k_2^\top \delta^\top k_1^\top k_1 \delta k_2
\;=\; k_2^\top \delta^2 k_2
$$
There are certainly variations depending on whether one wants to consider complex matrices, unitary versus orthogonal, and so on.
From such an expression, visibly, the singular values (EDIT: if this is truly what is intended) of the square are the squares of the singular values. For $SL_2(\mathbb R)$, the two singular values must be $t$ and $t^{-1}$.  Whichever is larger, when squared, is the larger of the two squares.  Since the maximum of $t$ and $t^{-1}$ is always at least $1$, the claim follows.
(Thanks to people trying to improve my answer, etc.)
EDIT-EDIT: (as @Jeppe Stig Nielsen notes in a comment), indeed, the operator norm is the square root of the largest eigenvalue of $A^\top A$, etc. Which is not so much related to the eigenvalues of $A$ itself, but to its "singular values" in an Iwasawa decomposition (see above). Perhaps this observation will resolve some confusion. (Believe me, these seemingly-slightly-different versions of similar things confounded me for a long time... :)
