Yes, the free product of the free group of rank $n$ and the free group of rank $m$ is isomorphic to the free group of rank $n+m$. Note also that if $G$ is free of rank $2$, and two elements $a$ and $b$ generate $G$, then $a$ and $b$ must freely generate $G$. One way to see this is to invoke the fact that a free group of finite rank is Hopfian. Another is to assume there is a nontrivial reduced word in $a$ and $b$ equal to the identity, take a free generating set $x$ and $y$, express $a$ and $b$ in terms of $x$ and $y$, and replace them in the nontrivial word expressing the identity; this will yield a nontrivial word in $x$ and $y$ equal to the identity (some work needs to be done here, of course), contradicting the choice of $x$ and $y$ as a free generating set. And there are other ways, of course.
An easy way to get this is as a corollary to the fact that the free group functor is the left adjoint of the underlying set functor. That is, for every group $G$ and every set $X$,
$$\mathcal{G}roup(\mathbf{F}(X),G) \longleftrightarrow \mathcal{S}et(X,\mathbf{U}(G)),$$
where $\mathbf{F}(X)$ is the free group on the set $X$ and $\mathbf{U}(G)$ is the underlying set of the group $G$.
Because the free group functor is a left adjoint, it sends coproducts to coproducts. That is, the coproduct of two free groups $F(X)$ and $F(Y)$ in the category of groups is the free group on the coproduct of $X$ and $Y$ in the category of sets. The coproduct in the category of groups is the free product, and the coproduct in sets is the disjoint union. Therefore, there is a natural isomorphism
$$F(X\amalg Y) \cong F(X)*F(Y).$$
You can also prove it directly from the universal property: the universal property of the free group on $X\amalg Y$ (the disjoint union) is that for every set-theoretic map $f\colon X\amalg Y\to G$ to a group $G$, there is a unique group homomorphism from $F(X\amalg Y)\to G$ that extends $f$. On the other hand, the universal property of the free product $F(X)*F(Y)$ is that for every pair of group homomorphisms $\varphi\colon F(X)\to G$ and $\psi\colon F(Y)\to G$, there is a unique group homomorphism $\Psi\colon F(X)*F(Y)\to G$ such that $\Psi\circ i_{F(X)}=\varphi$ and $\Psi\circ i_{F(Y)}=\psi$, where $i_{F(X)}\colon F(X)\to F(X)*F(Y)$ and $i_{F(Y)}\colon F(Y)\to F(X)*F(Y)$ are the canonical inclusions.
A set-theoretic map $f\colon X\amalg Y\to G$ is equivalent to a pair of maps $g\colon X\to G$ and $h\colon Y\to G$; the map $g\colon X\to G$ induceds a map $\varphi\colon F(X)\to G$, while the map $h\colon Y\to G$ induces a map $\psi\colon F(Y)\to G$, which in turn induces a map $F(X)*F(Y)\to G$; it is now straightforward to verify that this map extends $f$, and that it is unique, so that $F(X)*F(Y)$ has the universal property of $F(X\amalg Y)$, and therefore they are isomorphic.
Or: the inclusions $X\to F(X)\to F(X)*F(Y)$ and $Y\to F(Y)\to F(X)*F(Y)$ induce an inclusion $X\amalg Y\to F(X)*F(Y)$, which in turn induces a morphism $F(X\amalg Y)\to F(X)*F(Y)$. Conversely, the map $X\to F(X\amalg Y)$ induces a map $F(X)\to F(X\amalg Y)$, and $Y\to F(X\amalg Y)$ induces a map $F(Y)\to F(X\amalg Y)$, and these two maps together induce a map $F(X)*F(Y)\to F(X\amalg Y)$. It is now easy to verify that the induces maps $F(X\amalg Y)\to F(X)*F(Y)$ and $F(X)*F(Y)\to F(X\amalg Y)$ are inverses of each, in the usual abstract nonsense argument about their compositions having the same universal property as the corresponding identity.
P.S. The equality $F(a,b)=\langle a\rangle*\langle b\rangle$ presumably just means that there is a unique isomorphism between the two objects that maps $\{a\}$ and $\{b\}$ to themselves as the identity; this is a consequence of their respective universal properties/adjointness of free group construction.
We want to prove by induction on $n$ that every subgroup of an abelian group generated by $n$ elements is finitely-generated. If $n=1$, it is clearly true. From now on, suppose it is true for $n$.
Let $G= \langle a_1, \dots, a_{n+1} \rangle$ be an abelian group, $H \leq G$ be a subgroup and $\rho : G \to G/ \langle a_{n+1} \rangle$ be the quotient map. Notice that the group $G/ \langle a_{n+1} \rangle$ is generated by $\{ \rho(a_1) , \ldots, \rho(a_n) \}$, so by our induction hypothesis, the subgroup $\overline{H}:= \rho(H)$ is finitely generated: let $X=\{h_1,\dots, h_m\} \subset H$ be such that $\rho(X)$ generates $\overline{H}$.
On the other hand $H \cap \langle a_{n+1} \rangle$ is a cyclic group, say generated by $h_{m+1} \in H$. Now, we want to prove that $Y= \{h_1, \dots, h_m,h_{m+1} \}$ generates $H$.
Let $h \in H$. Of course, there exists a word $w \in \langle h_1,\dots, h_m \rangle$ such that $\rho(w)=\rho(h)$. Therefore, $h=w+k$ for some $k \in \mathrm{ker}(\rho)= \langle a_{n+1} \rangle$. Furthermore, $k=h-w \in H$ so $k=p \cdot h_{m+1}$ for some $p \in \mathbb{Z}$ since $\langle a_{n+1} \rangle \cap H = \langle h_{m+1} \rangle$. Finally, $$h=w+p \cdot h_{m+1} \in \langle h_1, \ldots, h_m,h_{m+1} \rangle = \langle Y \rangle,$$
so $Y$ generates $H$.
Best Answer
The following well-known equality is related to your question ($H$ and $K$ are arbitrary finite subgroups of a group $G$). $$|HK|=\frac{|H|\cdot |K|}{|H\cap K|}$$ See here for a proof. Note that $HK$ is not necessarily a subgroup, so does not always equal $\langle H, K\rangle$. On the other hand, if $HK$ is a subgroup then your result holds. Note that $HK$ is a subgroup if one of $H$ or $K$ is normal, and proving this is a nice exercise.
If $HK$ is a subgroup and $H\cap K=1$ then $HK$ is called the Zappa-Szep product of $H$ and $K$, written $H\bowtie K$. This is a generalisation of semidirect products (when either $H$ or $K$ is normal) and direct products (when both $H$ and $K$ are normal). See this question for more details. The answer mentions the specific case you are talking about here - the case when $H$ and $K$ are both finite cyclic. Such Zappa-Szep products were classified by Jesse Douglas, was one of two winners of the first Fields Medals (although he didn't win his medal for this...).