In general the right inverse of a linear bounded operator $T:X\to Y$ in Bartle-Graves theorem needs not to be linear, and there can be no linear bounded right inverse. In this case, any right inverse $S:Y\to X$ is everywhere not Fréchet differentiable nor Gateaux differentiable, because if $S$ is differentiable at $x_0$, by the chain rule $dS(x_0)$ is a bounded linear right inverse to $T$.

Note however that $\tilde S:Y\to X$ defined by $\tilde S(x):=\frac{\|x\|}2\Big( S\big(\frac x{\|x\|}\big)-S\big(-\frac x{\|x\|}\big) \Big)$ is also a right inverse to $T$, and it is
$1$-homogeneous, $S(tx)=tS(x)$. I'm not aware of other possible improvements one can do on $S$.

*There can be no linear bounded right inverse*: Precisely, a bounded linear operator $T:X\to Y$ between Banach spaces is a left, resp. right (bounded, linear) inverse, iff it is surjective and its kernel splits, resp. it is injective and its range is closed and splits.

In fact, if $T:X\to Y$ and $S:Y\to X$ are bounded linear operators such that $TS=1_Y$, then $Y=\ker T\oplus \text{ran}\, S,$ with linear projectors $I-ST=[T,S ]$ and $ST$.

boundedlinear surjective $G$ : If $Z$ is a Hilbert space, $U$ is a Banach space, $G : Z\to U$ is a bounded linear surjective operator, and $V:= (\ker G)^\perp$ with inclusion map $j:V\hookrightarrow Z$, then $G_{|V}:V\to U$ is invertible by the open mapping theorem and a bounded linear right inverse to $G$ is $j(G_{|V})^{-1}$. If $G$ is linear surjective between Hilbert spaces, and not assumed to be continuous, then the problem seems more difficult and I should think... $\endgroup$ – Pietro Majer Oct 31 at 21:04