There is no such function as can be seen from the work of Civin (1943) and Kerékjártó (1919).
Let me review what we need from the work of Civin (1943). Let $M$ be a compact manifold, and let $f$ be a continuous 2-to-1 function defined on $M$. For each $x\in M$, the preimage $f^{-1}(f(x))$ consists of $x$ and another point $s(x)\in M$. Let $K\subset M$ be the set of points where $s$ is continuous. For $x\in M$, let $t(x)=s(x)$ when $x\in K$, and let $t(x)=x$ when $x\not\in K$. Then $t:M\to M$ is a homeomorphism of order 2 (i.e. $t^2=1$); we shall call $t$ the homeomorphism associated with $f$. It is also known that $K$ is dense and open in $M$, and it is invariant under $s(x)$. Therefore, $F:=M\setminus K$ is a nowhere dense compact subset of $M$, which is invariant under $s(x)$, and the restriction of $f$ to $F$ is also 2-to-1. Note that $F$ is the set of fixed points of $t$.
Now assume that $f:S^2\to S^2$ is a continuous 2-to-1 function, and use the notations of the previous paragraph. By the theorem of Kerékjártó (1919), $t$ is conjugate within the group of homoeomorphisms of $S^2$ to a rotation of angle $\pi$ or a reflection. Therefore, we can assume without loss of generality (namely after composing $f$ from inside by a suitable homeomorphism of $S^2$), that $F$ is an antipodal pair of points or a great circle. In the first case, $f$ induces a homeomorphism from the annulus $S^2\setminus F$ divided by a rotation of angle $\pi$ (which is still an annulus) to the punctured sphere $S^2\setminus f(F)$. This is clearly absurd. Hence $F$ is a great circle, and $f$ induces a homeomorphism from either hemisphere connected component of $S^2\setminus F$ to $S^2\setminus f(F)$. Consider the restriction $g:=f_{\mid F}:F\to S^2$, which is a continuous 2-to-1 function, and the order two homeomorphism $u:F\to F$ associated with $g$. Similarly as before, $u$ is conjugate within the group of homeomorphisms of $F$ to a rotation of angle $\pi$ or a reflection. In either case, we can see that $f(F)$ is homeomorphic to $F$. Therefore, by the Jordan curve theorem, $S^2\setminus f(F)$ has two connected components, contradicting our earlier finding that it is homeomorphic to an open hemisphere.