The map $h : G/\ker f \to f(G)$ is essentially the same map as $f$, except that elements that were mapped to $0$ by $f$ are not taken into consideration anymore.
You can only do this when your subgroup is normal. As an example, consider the group $D_n = \langle r,s \, | \, r^n = s^2=1, srs = r^{-1} \rangle$. Take $n > 2$ for our example, and consider the two maps
$$
f_1 : D_n \to \mathbb Z/n \mathbb Z, \quad f_1(r^i s^j) = \overline i, \quad 0 \le i < n, \quad j \in \{0,1\}
$$
$$
f_2 : D_n \to \mathbb Z/2 \mathbb Z, \quad f_2(r^i s^j) = \overline j, \quad 0 \le i < n, \quad j \in \{0,1\}
$$
(The $\overline{i}$ means $i$ modulo $n$ ; I use the same notation for $j$ modulo $2$.) The first one is not even a group homomorphism ; we do not have the property that $f_1(s)f_1(r)f_1(s) = f_1(r^{-1})$ (check this ; this is where I use $n > 2$), which should have been true if $f_1$ was a group homomorphism. Intuitively, $f_1$ is an attempt to "mod out $s$". The reason why this attempt fails is explained by the fact that the subgroup $\langle s \rangle \subseteq D_n$ is not normal, i.e. $rsr^{-1} = r^2s \neq s$ when $n > 2$.
The second one, however, works just fine : it is a group homomorphism since
$$
f_2(r^{i_1} s^{j_1} r^{i_2} s^{j_2}) = f_2(r^{i_1 - i_2} s^{j_1 + j_2}) = \overline{j_1 + j_2} = \overline{j_1} + \overline{j_2}.
$$
This is illustrated by the fact that the subgroup $\langle r \rangle \subset D_n$ is normal, i.e. $r(r^i)r^{-1} = r^i$ and $s(r^i)s^{-1} = r^{-i}$.
So if we dismiss the information contained in a normal subgroup, we are fine ; this is because if $H \le G$ is a normal subgroup, then a product $gh$ where $g \in G$ and $h \in H$ will satisfy $gh = h'g$ where $h$ and $h'$ are elements of $H$. If you "don't read" $h$ and $h'$ in this equation, then you see $g$ on both sides. This is what being a normal subgroup means ; when we consider the elements of $H$ as behaving like the identity element in $G/H$, then we have a consistent group structure.
When you understand normal subgroups, you understand the projection map $\pi : G \to G/H$ for a normal subgroup $H$. The generalization $f : G \to K$ with $H = \ker f$ is not a big deal ; you can always replace $K$ by $f(G)$ (since $f(G)$ is a subgroup of $K$) and if you have read the proof of the first isomorphism theorem, you can always replace $f(G)$ by $G/H$, up to isomorphism. So your understanding of normal subgroups and the first isomorphism theorem reduces to understanding the projection map $\pi : G \to G/H$.
Hope that helps,
I assume you are having intuitive difficulties with the third statement of the theorem. Let me try and give an intuitive explanation. Every element of $SN$ is of the form $sn$ with $s \in S$ and $n \in N$. Now in $SN/N$ the $n$'s get 'killed' in the sense that in this group $\overline{sn}=\overline{s}$ for $s \in S$ and $n \in N$. However, we are not left with a group that is isomorphic with $S$, because if $s \in N$, that is if $s \in S \cap N$, then $s$ is also the identity in $SN/N$. So, we are left with $S$, but with the remaining part of $N$ completely filtered out, that is
$$\frac{SN}{N} \cong \frac{S}{S \cap N}$$
Best Answer
I understand the Theorem in the same way as you. The idea with a lot of these Algebra Theorems, where we factor an algebraic structure through a quotient, is to remove some undesirable part of the structure.
In this case, we want to get an isomorphism out of a surjective homomorphism, which is a much "nicer" map. So, we quotient out by the kernel, and as a result we have a map where only the zero element is sent to zero, which maintains surjectivity.
This sort of Theorem reappears frequently, and yours is the correct intuition.