$\newcommand{\im}{\operatorname{im}}$
Since the row is exact, we have $\im f \subset \ker g$. I must find a homomorphism $k:C\to D$ such that $k\circ g = h$. I have, at least, that $h\circ f = 0$, so $\im f \in \ker h$.
What's needed to create $k$ such that $k\circ g = h$ and it's a homomorphism? At least $g$ is homomorphism, so $k$ must be too, but what more?
Does that $0$ in the end helps? I think that, since the sequence is exact and that ons is the trivial homomorphism, it should help in something, right?
For uniqueness, of course I'd suppose $k\neq k'$ but I can't see what enforces them to be unique.
This boils down to the universal property of quotients:
The zero at the end does help, as it tells you that $\operatorname{Im}(g)=C$ and so $g$ is surjective. Now, by the first isomorphism theorem and exactness, you have an isomorphism:
$$\bar{g}:B/\operatorname{Im}(f)\to C,\quad \bar{g}(x+\operatorname{Im}(f)) = g(x).$$
That is, $\bar{g}\circ\pi=g$. Now, since $h\circ f=0$, we have $\operatorname{Im}(f)\subseteq\ker{h}$, and thus by the universal property there exists a unique map $\bar{h}:B/\operatorname{Im}(f)\to D$ such that $\bar{h}\circ\pi=h$. The composition $$k:C\xrightarrow{\ \bar{g}^{-1} \ }B/\operatorname{Im}(f)\xrightarrow{\ \bar{h} \ }D$$ is the map you need. I'll leave it to you to verify that $k\circ g=h$.
Uniqueness follows from the universal property as well: If $k':C\to D$ is any other map such that $k'\circ g =h$, then: $$h = k'\circ g = k'\circ\bar{g}\circ\pi,$$ but $\bar{h}$ is the unique map that $h$ factors through, and so we must have $k'\circ\bar{g}=\bar{h}$. This gives $k'=k$.