The gist of it : If you act two H gates on the qubit you will cancel the rotations thereby obtaining the initial qubit.
If you only pass one H gate then you have a superposition with a probability amplitude for each quantum state basis vector.
Longer explanation:
Just to illustrate this better I developed the first circuit with $\uparrow$:

Which will show the following state population distribution:

If we add $X$ gate to the circuit to bit flip $\uparrow$ into $\downarrow$, then we obtain the following:

With a similar state population measured counts:

Keep in mind the measurement number for this circuit is of 1000 times, so if you dial up this value you can find the distributions to get really close to each other.
Anyway, the idea I want to hammer home with the $H$ gates is that, it's an unitary, so $HH = H^2 = I$. This is why if you were to operate two of them in sequence, you will get back the original qubit for the $\uparrow$ state:

And for the $\downarrow$ state:

Now, when you have time later on and if you are curious:
We start by defining this as the Clifford Group (of Gates):
\begin{eqnarray}
G &=& \pm \Big \{ I, X, Y, Z \Big \}
\end{eqnarray}
This group is defined by the following properties:
- Each $\mathbf{M} \in G$ is unitary such that $M^\dagger = M^{-1}$.
- For each element $\mathbf{M} \in G, M^2 = \pm I$.
- If $\mathbf{M}^2 = I$, $\mathbf{M}$ is hermitian; otherwise, $\mathbf{M}$ is anti-hermitian.
- $\forall \, \mathbf{M_i}, \mathbf{M_j} \in G$, their products either commute or anti-commute, $\mathbf{M_i}\mathbf{M_j} = \pm \mathbf{M_j}\mathbf{M_i}$.
It turns out one can represent $H = \frac{1}{\sqrt{2}} \Big ( X + Z \Big )$ operation. Thus it falls under these rules.
This stuff is really, really, powerful stuff!