You are incorrect in your calculation of $\pi_1(U_g)$; it's free on $2g+1$ generators. What you need now is that the map $U_g \to U_{g+1}$ induces the map $F_{2g+1} \to F_{2g+3}$ sending the $i$th generator of the first to the $i$th generator of the second. Now we have...
If a space $X$ is the union of a directed set of subspaces $X_\alpha$ ordered by inclusion with the property that each compact set in $X$ is contained in some $X_\alpha$, and each $X_\alpha$ contains a given point $x_0$, then the natural map $\varinjlim \pi_1(X_\alpha,x_0) \to \pi_1(X,x_0)$ is an isomorphism.
Surjectivity follows from the compactness hypothesis; the image of a representing map $f: S^1 \to X$ of an element of $\pi_1(X,x_0)$ lies in some $X_\alpha$, so is in the image of some $f' \in \varinjlim\pi_1(X_\alpha, x_0)$. Suppose $f \in \varinjlim \pi_1(X_\alpha, x_0)$ maps to zero; then we have a null-homotopy $f_t: S^1 \times I \to X$. Since the image of this is compact, this defines a null-homotopy $f_t: S^1 \times I \to X_\alpha$ for some $\alpha$, so $f$ is zero in $\varinjlim \pi_1(X_\alpha, x_0)$ as well.
(This fact, and its proof, is an adaptation of Proposition 3.33 in Hatcher's algebraic topology book.)
Taking the limit of $\pi_1(U_g,x_0)$ gives us that $\pi_1(U) \cong F_\infty$, the free group on countably many generators. Similarly for $\pi_1(V,x_0)$.
Now $U \cap V \cong S^1$, and the inclusion $S^1 \to U, S^1 \to V$ induces the map $F_1 \to F_\infty$ sending the generator of the first to a chosen generator of the second. Thus by the van Kampen theorem, $$\pi_1(U \cup V, x_0) \cong \langle u_1, v_1, u_2, v_2, \dots \, \mid \, u_1 = v_1\rangle,$$
giving $\pi_1(X,x_0) = \pi_1(U \cup V, x_0) \cong F_\infty$ as desired.
Best Answer
Here's a proof that avoids covering space arguments, using a bit of differential topology, and using some surface topology arguments including the (smooth) Schönflies Theorem.
One can cut $S$ along a simple closed curve $\gamma$ (representing $[a,b]=[c,d]^{-1}$) into two compact subsurfaces $S_0,S_1$ with boundary, each a one-holed torus, each with fundamental group free of rank 2, so that the inclusion map $S_0 \hookrightarrow S$ takes a free basis of $\pi_1(S_0)$ to the two group elements $a,b$, so that the other inclusion map $S_1 \hookrightarrow S$ takes a free basis of $\pi_1(S_1)$ to $c,d$, and so that $\gamma = \partial S_0 = \partial S_1$ represents infinite order elements of $\pi_1(S_0)$ and of $\pi_1(S_1)$, namely $[a,b]$ and $[c,d]^{-1}$ respectively.
It suffices to prove that the inclusion map $S_0 \hookrightarrow S$ induces an injection on fundamental groups.
Since the inclusion of $\text{interior}(S_0)$ into $S_0$ is a homotopy equivalence, it suffices to prove that $\text{interior}(S_0) \hookrightarrow S$ induces an injection on fundamental groups. And for this, it suffices to prove that for every continuous closed curve $f : S^1 \to \text{interior}(S_0)$, if $f$ extends to a continuous function $F : D^2 \to S$ then $f$ also extends to a continuous function $F_0 : D^2 \to \text{interior}(S)$. By approximation methods in differential topology, we may assume that $f$ and $F$ are both smooth.
Now we apply transversality methods of differential topology. We know that $F^{-1}(\gamma) \subset \text{interior}(D^2)$, and so (applying perturbation methods in differential topology) we may assume that the map $F$ is tranverse to the curve $\gamma$. The subset $F^{-1}(\gamma) \subset \text{interior}(D^2)$ is therefore a pairwise disjoint collection of smooth, simple closed curves in the interior of $D^2$. Let $n$ be the number of those curves, denoted $C_1,...,C_n$.
By applying the Schönflies Theorem, each $C_i$ subdivides $D^2$ into two pieces: $\text{inside}(C_i)$ is diffeomorphic to $D^2$ with boundary $C_i$; and $\text{outside}(C_i)$ is diffeomorphic to an annulus with boundary $C_i \cup S^1$. Pick a collar neighborhood of $C_i$, parameterized as $C_i \times [-1,+1] \to D^2$ so that $C_n \times \{0\} \to C_i$ and $C_i \times \{+1\}$ is a simple closed curve $C'_i \subset \text{outside}(C_i)$.
The proof now proceeds by induction on $n$. The relation $C_i \subset \text{inside}(C_j)$ is a partial ordering on the finite set $\{C_1,\ldots,C_n\}$, and so there is a minimal element. Let's permute the notation so that $C_n$ is a minimal element, hence $\text{inside}(C_n)$ contains no other $C_i$. It follows that $F(\text{inside}(C_n)) \subset S_j$ for $j=0$ or $1$, and therefore $F | C_n $ is null homotopic in $S_j$. But $F | C_n \subset \gamma$ and $\gamma$ represents an infinite order element of $\pi_1(S_j)$, and so it follows that $F | C_n$ is null homotopic in the circle $\gamma$ itself. From this it follows that $F \mid C'_n$ is null homotopic in $\text{interior}(S_k)$ (namely in the complement of $S_j$, i.e. $k=1-j$). We can therefore redefine $F \mid \text{inside}(C'_n)$ to be a null homotopy of $F \mid C'_n$ in $\text{interior}(S_k)$. This redefinition operation removes $C_n$ from $F^{-1}(\gamma)$ with affecting the remainder of $F^{-1}(\gamma)$, thus reducing the number of components of $F^{-1}(\gamma)$, and completing the induction step.