Known-plaintext attack on 1-round Feistel networks Consider a simple, 1-round Feistel network.

Because the ciphertext is essentially $(L_1,R_1)$, if one knows the plaintext $(L_0,R_0)$, the only unknown left is $K$.

However round functions in Feistel networks are not necessarily invertible. For example, one could in principle use a cryptographic hash as a round function.

The Feistel network equation would become:

$R_1=L_0 \oplus \operatorname{sha2}(R_0,k_1)\\L_1=R_0$

One could XOR $R_1$ (known ciphertext) and $L_0$ (known plaintext), thus obtaining $\operatorname{sha2}(R_0,k_1)$, but this would not yield the key unless one uses brute force.

Clearly brute force is always possible - and there are other issues with a 1-round Feistel network, for example it leaks half of the plaintext - but I was wondering.

When one talks about KPA against 1-round Feistel networks, is it implicit that brute force can be used in the end to obtain $k$, for non-invertible round functions?
Regardless of the specific $f$ function, you already observe that you can obtain
$$
R_1\oplus L_0=f(R_0,k_1)
$$
where $k_1$ the first round subkey is your target.

But $R_0$ is also a part of the known plaintext. So you have a system of equations for the bits of $k_1$ which you can systematically attempt to solve. For a normal Feistel function $f$ (which is one to one or the cipher wouldn't be invertible) this is straightforward.

Exactly how $sha_2(R_0,k_1)$ is defined plays a part here: 

Possibilities include $sha_2(R_0||k_1),$  or $sha_2(R_0 \oplus k_1)$ (possibly with padding or repeating of $k_1$ to make it the same length as $R_0.$).

In any case you can try to use the $sha_2$ equations, noting that the effective input length is essentially at most the length of $R_0$ plus the length of $k_1$ and hence a strong hash function is most likely one to one for such inputs. Using the random oracle model, a randomly chosen function from an input of length, say, 128 bits, to the full digest size for sha_2 of 512 bits is one to one with overwhelming probability (see birthday paradox, addressed in many answers to questions on crypto stackexchange).