Is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?

As for your first question:
This is a purely mathematical tactic in order to obtain an (approximate) equation for $R$. The expansion of first order on the RHS is motivated by this fact, i.e. to bring $R$ alone "in the surface". The reason why the LHS is subjected to a second-order expansion is in order for something to be left (the variance term). Higher-order expansions of the RHS can certainly be applied.

As for your second question, you are making a mistake. The center of expansion for the RHS expansion is the same as that for the LHS expansion, namely $w-E(x)$ (or equivalently, around $R=0$). It is meaningless (and it fails) to expand a function around its exact argument. Specifically we have

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] = u\left(w+E(x)\right) - u'\cdot R$$

Finally, why not consider a second-order expansion on the RHS? We would then get

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] \\+\frac 12 u''\cdot [(w+E(x)-R)-(w+E(x))]^2 $$

$$= u\left(w+E(x)\right) - u'\cdot R + \frac 12 u''\cdot R^2$$

Then we would obtain a quadratic polynomial in $R$,

$$\frac 12 u''\cdot R^2 - u'\cdot R - \frac 12 u''\sigma^2_x = 0$$

$$\ R^2 - \frac {2u'}{u''}\cdot R - \sigma^2_x = 0$$

This has roots

$$R_1,R_2 = \frac {(2u'/u'') \pm \sqrt{(2u'/u'')^2+4\sigma^2_x}}{2}$$

$$\implies R = \frac{u'}{u''} + \sqrt{\left( \frac{u'}{u''}\right)^2+\sigma^2_x}$$

You can totally validly use this expression for $R$, but I guess you understand why the simpler one is used instead.

Is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?

As for your first question:
This is a purely mathematical tactic in order to obtain an (approximate) equation for $R$. The expansion of first order on the RHS is motivated by this fact, i.e. to bring $R$ alone "in the surface". The reason why the LHS is subjected to a second-order expansion is in order for something to be left (the variance term). Higher-order expansions of the LHS can certainly be applied.

As for your second question, you are making a mistake. The center of expansion for the RHS expansion is the same as that for the LHS expansion, namely $w-E(x)$ (or equivalently, around $R=0$). It is meaningless (and it fails) to expand a function around its exact argument. Specifically we have

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] = u\left(w+E(x)\right) - u'\cdot R$$

Finally, why not consider a second-order expansion on the RHS? We would then get

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] \\+\frac 12 u''\cdot [(w+E(x)-R)-(w+E(x))]^2 $$

$$= u\left(w+E(x)\right) - u'\cdot R + \frac 12 u''\cdot R^2$$

Then we would obtain a quadratic polynomial in $R$,

$$\frac 12 u''\cdot R^2 - u'\cdot R - \frac 12 u''\sigma^2_x = 0$$

$$\ R^2 - \frac {2u'}{u''}\cdot R - \sigma^2_x = 0$$

This has roots

$$R_1,R_2 = \frac {(2u'/u'') \pm \sqrt{(2u'/u'')^2+4\sigma^2_x}}{2}$$

$$\implies R = \frac{u'}{u''} + \sqrt{\left( \frac{u'}{u''}\right)^2+\sigma^2_x}$$

You can totally validly use this expression for $R$, but I guess you understand why the simpler one is used instead.