We can do this by considering the "universal" definition of $N$: $N$ has the following properties:
- $H\subseteq N$;
- $N\triangleleft G$; and
- If $K$ is any subgroup such that $H\subseteq K$ and $K\triangleleft G$, then $N\subseteq K$.
Since $\mathrm{ker}(f)\triangleleft G$, and we are assuming that $H\subseteq \mathrm{ker}(f)$, property 3 implies immediately that $N\subseteq \mathrm{ker}(f)$.
We can also do this by considering the top-down construction of $N$: $N$ is the intersection of all normal subgroups of $G$ that contains $H$: $N = \bigcap_{H\subseteq K\triangleleft G} K.$ Then since $H\subseteq \mathrm{ker}(f)\triangleleft G$, it follows that $N\subseteq\mathrm{ker}(f)$.
Or we can do this by considering the bottom-up construction of $N$; but here you are somewhat incorrect above. You write that "every element of $N$ can be written as a product containing elements of $H$", and this is not quite good enough. The correct statement is that $N$ consists exactly of products of conjugates of elements of $H$: $N = \Bigl\{ (g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigm| g_i\in G, h_i\in H, m\in\mathbb{N}\Bigr\}.$ Given such an element, if we apply $f$ to it we get: $\begin{align*} f\Bigl((g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigr) &= f(g_1)f(h_1)f(g_1)^{-1}\cdots f(g_m)f(h_m)f(g_m)^{-1}\\ &= f(g_1)1f(g_1)^{-1}\cdots f(g_m)1f(g_m)^{-1} \quad\text{(since }H\subseteq \mathrm{ker}(f)\text{)}\\ &= 1, \end{align*}$ so for every $n\in N$ we have $f(n)=1$; hence, $N\subseteq \mathrm{ker}(f)$.