Let me try the following argument.

First, as you explain one can reduce the general case to the case when $X$ is a curve $C$, possibly singular and reducible. This is because the preimage of $g(\mathbb{P}^1)$ under the quotient morphism $\phi: X^n \to S^n(X)$ has dimension dimension one (or zero if $g$ is constant), and one can find a closed reduced curve $C \subset X$ such that the image $g(\mathbb{P}^1)$ is contained in $S^n(C) \subset S^n(X)$, and thus $g$ will factor through $S^n(C)$.

We need to show that $g: \mathbb{P}^1 \to S^n(C)$ factors through $S^n(\tilde{C}) \to S^n(C)$ where $\tilde{C} \to C$ is the normalization. We can reduce to the case when $g: \mathbb{P}^1 \to S^n(C)$ represents divisors on $C$ with no fixed components. Indeed, if $x \in C$ is a point in the support of every divisor in the family, the morphism $g$ will factor as a composition $\mathbb{P}^1 \to S^{n-1}(C) \to S^{n}(C)$, where the latter map maps an effective divisor $D$ on $C$ to $D + x$. This way we can get rid of all fixed components while decreasing $n$.

Finally, if $g$ has no fixed components, then for every point $x \in C$ the set of $t \in \mathbb{P}^1$ such that $x$ is contained in the support of $g(t)$ is finite. In particular, if we let $U \subset \mathbb{P}^1$ to denote the subset parametrizing divisors supported at the smooth locus of $C$, the set $U$ will be open and non-empty. Furthermore it is clear that we have a morphism $U \to S^n(\tilde{C})$ lifting $g|_U$, because normalization is as isomoprhism away from the singular locus of $C$. By the valuative criterion of properness, we can extend $U \to S^n(\tilde{C})$ to a morphism $\tilde{g}: \mathbb{P}^1 \to S^n(\tilde{C})$, and it will automatically be a lifting of $g$ (since it is such on an open dense subset).