If $R[s], R[t]$ are finitely generated as $R$-modules, the so is $R[s + t]$. 
Let $S \supset R$ as rings with $1 \in R$.  Suppose that $s, t \in S$ and that the subrings $R[s], R[t]$ are finitely generated by $\{1, s, \dots, s^k\}$ and $\{1, t, \dots, t^m \}$.  Then $R[s + t]$ is also finitely generated.

Let $g$ be a minimal degree monic polynomial for $s$ over $R$.
I'm thinking let $f(X) = X  + t. \ $  Then $s + t = f(s). \ $  So  $f(X)^n = q(X) g(X) + r(X)$ by the division algo, where $r(X) = 0$ or $\deg{r} \lt \deg{g}$.  Then $f(s)^n = (s+t)^n = 0 + r(s), \ $ for some $r(X) \in R[t][X]$.
Now I'm very confused, but this looks like the right direction. Please work within it if you can, but all answers accepted.  Thanks.

Let $R[s]$ be fin-gen with monic $f$ and $R[t]$ fin-gen with monic $g$.
$(X + t)^k, k \geq 1$, is a polynomial in $R[t][X]$ .  Since $s$ is integral over $R$ with $f$ it's also integral over $R[t]$ with $f$ since $R \subset R[t]$.  Then $(X + t)^k = q(X) f(X) + r(X)$ for some $r = 0$ or $\deg r \lt \deg f$.  This means that if $h(X) = (X + t)^k$, then $h(s) = (s + t)^k$ and $h(s) = r(s)$ for some $r$.  So we have that $p \in R[s+t]$ can be expressed as $p = p_0 + p_1 (s+t) + \dots p_n (s+t)^n = p_0 + p_1 r_1(s, t) + \dots + p_n r_n(s,t)$, each $r_i$ having degree $\leq \deg f + \deg g$, and each of which is a polynomial in $\{s^{i}t^{j} : 0 \leq i,j, i+j \leq \deg f + \deg g \} $.  This set is finite and clearly generates $R[s + t]$ over $R$ as an $R$-module.
QED
