$\newcommand{\Z}{\mathbb Z}$ Let $K\leq F_2 = \langle a,b\rangle$ be the kernel of the map $\Phi: F_2 \to \Z$ sending $a$ to $2$ and $b$ to $3$. Draw a cover of $S^1 \vee S^1$ whose fundamental group maps isomorphically to $K$ under the homomorphism induced by the covering map.
Let $\sigma \in K\leq F_2$. Since $K=\ker \Phi$, we must have that $\Phi(\sigma)=0$ for any element $\sigma$ of $K$. In particular, $K$ will contain words in $F_2$ on $a$ and $b$ such that the total (not necessarily contiguous) exponents of $a$ will be $3n$ and the total exponents of $b$ will be $-2n$ for some $n\in \Z$. Hence, we will find that $K$ is generated by the $10$ terms given below, with a covering of $S^1 \vee S^1$ drawn whose fundamental group maps isomorphically to $K$ under the homomorphism induced by the covering map.
\begin{align*} K = &\langle a^3b^{-2},a^2 b^{-2}a, ab^{-2}a^2, b^{-2}a^3, b^{-1}a^3 b^{-1}, b^{-1}a^2 b^{-1}a,\\ &b^{-1}ab^{-1}a^2, ab^{-1}ab^{-1}a,a^2 b^{-1}ab^{-1},ab^{-1}a^2 b^{-1} \rangle. \end{align*}
Is this correct?
This all looks correct except for your description of a generating set for $K$. I think you were fooled by your (otherwise very nice) picture, forgetting that the picture of $K$ repeats infinitely going off to the left and to the right. The generators that you have written do not generate all of $\pi_1 K$: it looks like the portion of $\pi_1 K$ that they generate is "approximately" the portion that is visible in your picture. Roughly speaking, in order to get a full generating set for $\pi_1 K$ you should write down a few basic generators and then write down all of their conjugates as well. So, for instance, given that your generating set for $\pi_1 K$ includes $a^{3}b^{-2}$, it should also include $a^{42} a^3 b^{-2} a^{-42}$.
In order to get a free basis for $\pi_1 K$, you do the thing one learns to do for the fundamental group of any graph: first pick a base point and a maximal tree, then enumerate the edges that are not in the maximal tree, and then write down the appropriate basis element for each edge. In this example, for the maximal tree you could use the upper line of $a$ edges union each negative slope $b$ edge. Each free basis element would therefore have one of two forms: $a^n b a b^{-1} a^{-n-1}$ associated to each $a$ edge along the lower line; and $a^n b^{-2} a^{-n+3}$ associated to each positive slope $b$ edge.
There is a general theorem about any normal subgroup $K < F$ of any free group $F$ which says that if $K$ is nontrivial and of infinite index then $K$ is not finitely generated.