This is again false. The geometric interpretation is as follows: write $Y = \operatorname{Spec} S$ and $X_i = \operatorname{Spec} R_i$. Given étale morphisms $f_1 \colon Y \to X_1$ and $f_2 \colon Y \to X_2$ of affine schemes, the image factorisation $$R_1 \underset k\otimes R_2 \twoheadrightarrow R \hookrightarrow S$$ corresponds geometrically to taking the (scheme-theoretic) image $Y \to X \hookrightarrow X_1 \times X_2$ of the product map $f \colon Y \to X_1 \times X_2$. If $Y$ is smooth, then so are the $X_i$, but $X$ need not be smooth. If $f$ is surjective, it cannot be flat when $X$ is singular [Tags 07NG and 00HQ].
We can turn this around to make a counterexample: start with a map $f \colon Y \to X_1 \times X_2$ of smooth schemes such that the image is singular but both projections $f_i \colon Y \to X_i$ are étale.
Example. This is the typical picture of a plane nodal curve: let $Y_0 = X_1 = X_2 = \mathbf A^1$, and consider the morphism $f \colon Y_0 \to X_1 \times X_2$ given by $t \mapsto (t^2-1,t^3-t)$. The scheme-theoretic image is the nodal curve $X_0 = V(y^2 - x^2(x+1))$, and the map $Y_0 \to X_0$ is surjective.
The projections $f_i \colon Y_0 \to X_i$ are not étale, but they become so after removing the points $0$ and $\pm\tfrac{\sqrt 3}{3}$ from $Y_0$. Let $Y \subseteq Y_0$ (resp. $X \subseteq X_0$) be the complement of these points (resp. their images in $X$). This gives the desired counterexample, since we did not remove the singular point $(0,0) = f(-1) = f(1)$ from $X$.
(This parametrisation is explained on this page of Donu Arapura, which is literally the first result that my favourite search engine produces when I type 'nodal curve'.)
Translated back into algebra, this means $S = k\bigl[t,\tfrac{1}{t},\tfrac{1}{3t^2-1}\bigr]$, with $R_1 = k[t^2-1] \subseteq S$ and $R_2 = k[t^3-t] \subseteq S$.