Your question is essentially answered in a math.SE post by Asaf Karagila, but I think it is worth spelling out a subtle point. The axiom of multiple choice (or MC for short) says:

For every set $X$ of nonempty sets, there exists a function $f$ on $X$ such that for every $x\in X$, $f(x)$ is a finite nonempty subset of $x$.

In Lemma 2 of Some theorems on vector spaces and the axiom of choice (Fund. Math. 54 (1964), 95–107), M. N. Bleicher showed that MC is implied by the following statement:

For some field $F$, every subspace of a vector space over $F$ has a complementary subspace.

In fact, Bleicher's results combined with a result of M. C. Armbrust (An algebraic equivalent of a multiple choice axiom, Fund. Math. 74 (1972), 145–146) imply that the above statement is equivalent to MC.

The subtle point is that the above equivalence can be proved not just in ZF, but in some other set theories such as ZFU (a variant of ZF with "urelements" or "atoms"). So this equivalence is "robust" in some sense. On the other hand, ZF proves the equivalence of MC with the axiom of choice (for a non-paywalled proof, see Theorem 2.18 in Kevin Barnum's note, The axiom of choice and its implications), but this equivalence is "delicate" because if you switch to a slightly different set theory, MC might be strictly weaker than AC.

So the answer to your question is that your "apparently weaker" statement is equivalent to AC if you work over ZF. But over some other closely related set theories, I'm not sure the answer is known. Some further information may be found in Paul Howard's paper, Bases, spanning sets, and the axiom of choice (Math. Logic Q. 53 (2007), 247–254), and Marianne Morillon's paper, Linear extenders and the axiom of choice (Comm. Math. Univ. Carol. 58 (2017), 419–434).