Normal generation of the kernel of a surjection of free groups

The answer is yes, and $n$ is an upper bound on the required number of normal generators. I don't know whether there is a better bound (such as $n-m$ perhaps)?

Let $x_i$ ($1 \le i \le n$) and $y_j$ $(1 \le j \le m)$ be free generators of $F_n$ and $F_m$, and choose elements $g_j \in F_n$ with $f(g_j) = y_j$ for $1 \le j \le m$.

Now for each $x_i$, we can write $f(x_i)$ as a word $w_i(y_j)$ in $y_j^{\pm 1}$, and $z_i := x_i^{-1} w_i(g_j) \in \ker f$ for $1 \le i \le n$.

I claim that $\ker f$ is normally generated by $\{ z_i : 1 \le i \le n \}$. Let $N = \langle z_i \rangle^{F_n}$, so $N \le K$, and we want to prove equality.

Now, since any word in the $x_i$ can be rewritten modulo $N$ as a word in the $w_i(g_j)$, we see that $F/N$ is generated by the elements $w_i(g_j)$, and hence also by $\{ g_j : 1 \le j \le m \}$. So $F/N$ is isomorphic to a quotient of $F_m$ and, since free groups of finite rank are known to be Hopfian (see here for example), we have $K=N$ as claimed.

Let me elaborate on Derek Holt's answer, providing some kind of generalization, where we see exactly how the bound appear. The general statement I propose is the following:

Let $p: G \twoheadrightarrow K$ be a split epimorphism between groups, $s: K \hookrightarrow G$ be a section of $p$, and $H$ be the kernel of $p$. If $(x_i)_{i \in I}$ is a generating family for $G$, then the family $(x_i \cdot sp(x_i)^{-1})_{i \in I}$ normally generates $H$.

Here is a proof: Let $N$ be the subgroup of $G$ normally generated by the $x_i \cdot sp(x_i)^{-1}$. These are elements of $H$, hence $N \subseteq H$. Thus $p$ factors through the quotient $\pi : G \twoheadrightarrow G/N$ : there is a unique $\bar p : G/N \rightarrow K$ such that $\bar p \pi = p$. Now, $\pi s$ is a section of $\bar p$, since $\bar p \pi s = ps = id_K$. This implies that $\pi s$ is injective. Moreover, $\pi s$ is surjective, by definition of $N$. Indeed, $G/N$ is generated by the $\pi(x_i)$, and $\pi(x_i) = \pi(sp(x_i)) = \pi s(p(x_i))$ : the image of $\pi s$ contains the $\pi(x_i)$, whence all of $G/N$. Thus, $\pi s$ is an isomorphism, whose inverse mapping is $\bar p$. As a consequence, $1 = \ker(\bar p) = H/N$, which means that $H = N$, and the statement is proved.

In order to see that the answer you seek is a particular case of this, you need to remark that a surjection onto a free group always splits : any choice of lifts of generators ($y_j \mapsto g_j$ in Derek Holt's answer) extends to a section of your epimorphism. Then the $w_i(g_j)$ in Derek Holt's answer are exactly my $sp(x_i)$, and you see that our answers are the same in this case.

And from the look of my answer, I am guessing that the best bound is indeed $n$, even in the case of free groups (but I do not have a good example at hand ; I need to think about that).