Frobenius splitting for an excellent, non $F$-finite, $F$-pure hypersurface Let $p$ be an odd prime. Let $k$ be a field of characteristic $p$ such that $[k:k^p]=\infty$ (i.e. $k$ is not $F$-finite ) . 
Also assume that $-1$ is not a square in $k$ . Consider the homogeneous polynomial $f(x,y)=x^2+y^2\in k[x,y]$ . Then $f$ is irreducible in $k[x,y]$ , hence $R=k[x,y]/(f)$ is an integral domain (of dimension $1$ , hence Cohen-Macaulay) . 
Moreover, $R$ is Excellent. (https://en.m.wikipedia.org/wiki/Excellent_ring) 
Also, $R$ is not $F$-finite (as $k$ is not). 
We also observe that by Fedder's criteria, $R$ is $F$-pure since $f^{p-1}\notin (x^p,y^p)$ .
Indeed, to see $f^{p-1}\notin (x^p,y^p)$, observe that $f^{p-1}=(x^2+y^2)^{p-1}=x^{p-1}y^{p-1}+\sum_{0\le j\le p-1 , j\ne \dfrac {p-1}2 } (x^2)^{j}(y^2)^{p-1-j}$. 
Now $x^{p-1}y^{p-1}\notin (x^p, y^p)$.
Moreover, either $j$ or $p-1-j$  is $\ge \dfrac{p-1}2+1=\dfrac {p+1}2 $ if $j\ne \dfrac {p-1}2$, hence $(x^2)^{j}(y^2)^{p-1-j}\in (x^p, y^p)$ when $j\ne \dfrac {p-1}2$. Thus $f^{p-1} \notin (x^p, y^p)$. 
My question is: When can we say that $k[x,y]/(x^2+y^2)$ is $F$-split ? 
 A: Your ring is Frobenius split. I also don't think you need the condition that $k$ does not contain a square root of $-1$.
Proof. Denoting by $\overline{k}$ the algebraic closure of $k$, we see that
$$\overline{R} := R \otimes_k \overline{k} \simeq \frac{\overline{k}[x,y]}{x^2+y^2}$$
is $F$-pure by Fedder's criterion, using the same computation you gave. Since $\overline{k}$ is algebraically closed, it is $F$-finite. Thus, $\overline{R}$ is also $F$-finite. Combining these two facts, we see that $\overline{R}$ is Frobenius split.
Now consider the commutative diagram
$$\require{AMScd}\begin{CD}
  R @>F_R>> F_{R*}R @>\phi>\exists?> R\\
  @VVV @VVV @VVV\\
  \overline{R} @>F_{\overline{R}}>> F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R}
\end{CD}\tag{1}\label{eq:basechange}$$
where the composition in the bottom row is the identity on $\overline{R}$ and the vertical maps are obtained by extending the field extension $k \subseteq \overline{k}$ by scalars along $k \to R$.
We want to show that the map $\phi$ exists such that the composition in the top row is the identity on $R$. Since $k \subseteq \overline{k}$ splits as a map of $k$-vector spaces, the map $R \to \overline{R}$ splits as a map of $R$-modules. Let $f\colon \overline{R} \to R$ be such a splitting. Then, defining $\phi$ to be the composition of the three arrows in the right square of the diagram
$$\begin{CD}
  R @>F_R>> F_{R*}R @>\phi>> R\\
  & @VVV @AAfA\\
  & & F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R}
\end{CD}$$
we see that the composition in the top row of the diagram \eqref{eq:basechange} is the identity on $R$. Thus, $R$ is Frobenius split. $\blacksquare$
I also want to point out that this is a special case of a result due to Rankeya Datta and myself. The statement for $F$-purity for complete local rings I believe first appeared in [Fedder, Lem. 1.2].
Theorem (Datta–M; see [M1, Thm. B.2.3]). Let $R$ be a ring essentially of finite type over a noetherian complete local ring $(A,\mathfrak{m})$ of prime characteristic $p > 0$. Then,


*

*$R$ is $F$-pure if and only if $R$ is Frobenius split; and

*$R$ is strongly $F$-regular if and only if $R$ is split $F$-regular.
Here, $R$ is strongly $F$-regular if every inclusion of modules is tightly closed, following [Hochster, Def. on p. 166], and $R$ is split $F$-regular if for every element $c$ avoiding every minimal prime of $R$, there exists an integer $e > 0$ such that the composition
$$R \overset{F^e_R}{\longrightarrow} F^e_{R*}R \xrightarrow{F^e_*(-\cdot c)} F^e_{R*}R$$
splits as a map of $R$-modules. Split $F$-regularity is the original definition for strong $F$-regularity in the $F$-finite case [Hochster–Huneke, Def. 5.1], although the terminology comes from [Datta–Smith, Def. 6.6.1].
The proof of the theorem uses the gamma construction of Hochster–Huneke [Hochster–Huneke, (6.7) and (6.11)] and [M2, Thm. 3.4], but the idea is to construct a diagram like that in \eqref{eq:basechange}, and use the fact (communicated to Hochster by Auslander) that pure maps from complete local rings split.
In your case, you can set $A$ to be the ground field $k$.
Finally, it is worth mentioning that the theorem does not hold for excellent rings in general: Rankeya and I found excellent regular rings, and even DVR's, that are not Frobenius split [Datta–M]. The rings we consider are Tate algebras and rings of convergent power series over non-Archimedean valued fields. The specific field we use was provided to us by Gabber, but one can also use fields constructed by Blaszczok and Kuhlmann.
