Proof:

Let $H$ be generated by $x_1, x_2,\dots, x_k$ and for each $j = 1, 2, \dots, k$ let $G_j$ be the infinite cyclic group with generator $a_j$, regarded as a subgroup of $F_k$. Applying the definition of the coproduct to the collection of group homomorphisms $f_j:G_j\longrightarrow H$ defined by

$$f_j(a_j) = x_j,\quad j = 1, 2\dots, k,$$

we get a group homomorphism $\phi:F_k\longrightarrow H$ such that $\phi(a_j) = f_j(a_j) = x_j$. It is clear that $\phi$ is surjective and the proof is complete.