Consider the following Lemma:
Let $K$ be a normal subgroup of $G$ and $T$ be a subgroup of $G/K$,then there exists a subgroup $H$ of $G$ such that $T=H/K$. (Of course, it is automatically implied that $H$ must contain $K$ or else $H/K$ makes no sense.)
(Actually it says that any subgroup of quotient group by $K$,has the form $H/K$ where $H$ is a subgroup of $G$.)
I am trying to proceed intuitively to get hold of the formal proof.
Quotienting by a normal subgroup essentially means consider the cosets induced by the subgroup and collapse each to a point. Now $T$ is a subgroup of $G/K$, so basically it is a collection of points from $G/K$ which are actually some cosets of $K$ in $G$ (reduced to a point).
So, let us expand the coset-points to the original parent points. I mean to say, we have collapsed cosets via the epimorphism $f(x)=xK ,x\in G$, so let us now expand the coset points, i.e., go reverse.
Consider $f^{-1}\{yK $ cosets of $G\mid yK \in T\}$, i.e., $f^{-1}(T)$, which is basically the union of all the cosets present in the set $T$. Now clearly this set contains $K$ as $K$ is in $T$ because $K=eK$ is the identity of $G/K$ and $T$ is a subgroup of $G/K$. One can easily verify this to be a subgroup of $G$ by the properties of homomorhism and since we have expanded coset points to get the set $f^{-1}(T)$, so collapsing it again will give the same coset point collection, i.e., $T$, so of course $T=H/K$.
Is my intuition correct?
Your intuition is correct -- if $H \leq G/K$, then we can write $H = \widetilde{H}/K$ where $\widetilde{H} \leq G$.
The idea is to set $\widetilde{H} = f^{-1}(H)$. It is routine to verify that this is a subgroup of $G$, and that $f(\widetilde{H}) = f(f^{-1}(H)) = H$.
Here is the requested picture for how I visualize what is happening. It is similar in spirit to the description you gave:
The bubbles over a point $x \in G/K$ are exactly the $g \in G$ such that $f(g) = x$. You'll notice these are all isomorphic, as they are exactly the cosets of $K$. Thus each bubble looks like a copy of $K$. Then a subgroup $H \leq G/K$, shown as the circled region, lifts to a subgroup in $G$ which looks like $H$ copies of $K$. This is the set of bubbles offset from the others. If we ignore the bubbles and just look at the 24 points in $G$ which lie above $H$, this is exactly $\widetilde{H}$, and this picture makes it clear that projecting these down to $G/K$ gives exactly $H$.
I hope this helps ^_^