Secret sharing scheme WITHOUT Shamir Secrete Sharing
XOR secret sharing used directly is unconditionally secure, if the randomly generated first share $S_1$ is a uniformly distributed vector in $\{0,1\}^{128}$. Let the secret be $X \in \{0,1\}^{128}$ and the shares be $S_2=X+S_1,$ and $S_1,$ where $+$ denotes bitwise XOR.

For a third party to learn the secret, they need to brute force $O(2^{128})$ guesses in the worst case.

Now you've "hidden" the secret $X$ by your $S_1'=TRUNC_{128}(SHA_{512}(X)).$

As stated in the comments I was wrong, thebirthday paradox does not directly apply.

An attacker brute forcing the share $S_2’=X+X_1$ at the same complexity will not learn your secret but its hash, so you’ve gained something.

However in the absence of side information, the original XOR scheme is as secure, unless the guesses for the secret can be tested by using it, say, as a password to unlock some system. In that use case, your modification is an improvement since it prevents this way of testing if the guess is correct.