Immediately above the surface, write
$$
\mathbf{E}=E_{\text{above}}\widehat{\mathbf{n}}+\mathbf{E}_\text{tangential,above}
$$
and immediately below the surface, write
$$
\mathbf{E}=E_{\text{below}}\widehat{\mathbf{n}}+\mathbf{E}_{\text{tangential,below}},
$$
where $\widehat{\mathbf{n}}$ is the unit surface normal pointing in the above direction, and both tangential components, by definitition, are orthogonal to $\widehat{\mathbf{n}}$. In the limit where the "pillbox" gets arbitrarily small, we can treat $E_{\text{above}}$ and $E_{\text{below}}$ as approximately constant in the pillbox, so that the flux is given by
$$
A(E_{\text{above}}-E_{\text{below}})+O(h),
$$
where $h$ is the height of the pillbox (this is Big $O$ notation). The minus sign arises because the unit normal to the surface below is $-\widehat{\mathbf{n}}$, not just $\widehat{\mathbf{n}}$. By Gauss's Law, this is equal to $\frac{A\sigma}{\varepsilon _0}$, and hence, in the limit
$$
\boxed{E_{\text{above}}-E_{\text{below}}=\frac{\sigma}{\varepsilon _0}}.
$$
Obviously, techincally speaking more careful limiting arguments are needed, but hopefully it is clear that this argument can be made completely precise. Let me know if you have any more questions.
To be fair, this is essentially exactly what Ron was getting at, but hopefully the increased detail made this more clear.
Decomposing the E-field: since this is a vector, it can be expressed equivalently in terms of any three basis vectors that span the $\mathbb{R}^3$ vector space (which is intuitively just the set of "arrows" in three dimensions). The standard basis choice is $\{\hat{x},\hat{y},\hat{z}\}$, but any three linearly independent vectors will do. We shall almost always want to choose orthonormal vectors (mutually perpendicular, all having length 1). In this case, being that we are near a surface of charge, we want basis vectors that are useful in this small local region: one unit vector normal to the surface (usually denoted $\hat{n}$), and two that are tangent to the surface at this particular point (call them $\hat{s}_1$ and $\hat{s}_2$). So to actually perform the decomposition given $\vec{E}=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}$, we would first need to find formulas for $\hat{n}$ and $\hat{s}_i$ in the xyz-basis also, then invert these formulas to solve for $\{\hat{x},\hat{y},\hat{z}\}$ in terms of $\{\hat{n},\hat{s}_i\}$, then simply plug these into the equation for $\vec{E}$ and regroup.
I am not sure how to answer "why" the normal component is discontinuous and the parallel (i.e. tangent) components are discontinuous. The explanation is given right there in Griffiths: it is due to Gauss' Law and the vanishing of the curl. Perhaps you could elaborate on what you are not understanding about it.
You seem to be severely confused when you talk about the "direction" changing of a particular component of the E-field. Strictly speaking, each component is just a number, so it doesn't have any direction at all, let alone one that can change. For instance, in standard Cartesian xyz-coordinates, the component $E_x$ just tells us the amount of E-field that points in the x-direction. The direction corresponding to each component is fixed by the unit vector it is paired with, and the total E-field is the vector sum of all the basis vectors weighted by the components in each direction. This is why we talk about the normal component, and the tangent components (note the plural, since there are two independent directions in the plane tangent to the charge surface at the point were are focussing on).
For continuity, I'm not sure where you're getting anything about $\mathbb{R}^2$ from. Continuity in this case just refers to each component as a function of a single variable (the distance from the surface). We are concerned with what happens above and below this surface, so imagine an axis that is normal to the surface, piercing it at the particular point we are interested in. Since this is a one-dimensional object, we can label points along it with a single variable which we'll call $n$ (pop quiz: what unit vector corresponds to this coordinate variable?). Since all the points on this axis are just points in space, the E-field is defined along the n-axis, and hence we can consider $\vec{E}(n)$, a vector function of a single variable. But "vector function" just means three regular functions, one for each component: $E_n(n), E_{||,1}(n),$ and $E_{||,2}(n)$. One normal component, and two tangent components, which despite the names are only actually normal/tangent AT the surface itself, it is just that we are still talking about the field in terms of the basis vectors we defined at the surface even though we are considering points which are above and below the surface. If we call $n=0$ where the normal axis intersects the surface, then all we are saying is that these three functions are (dis)continuous at $n=0$. Note that we can combine the two tangent components into a 2D vector $\vec{E}_{||}$, which is what Griffiths does.
The equation $$\Delta\vec{E}=\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0}\hat{n}$$ just summarizes everything Griffiths has just deduced about the three different components into a single vector equation. Remember that we're writing our vectors in terms of three orthonormal basis vectors, and due to the orthonormality you can find the components by dotting your vector with each basis unit (e.g. $\hat{x}\cdot\vec{E} = E_x$). This tells us that $\Delta\vec{E}$ has no tangent components (try dotting either $\hat{s}_i$ into it), i.e. the tangent components are continuous. Similarly, there is a difference in the normal component of the E-field given by $\sigma/\epsilon_0$.
Best Answer
You can assume $\nabla\cdot \vec{A} =0$.
Can you see that this is analogous to the no-monopole law of $\nabla \cdot \vec{B} =0$?
Therefore, just like the magnetic field, the component of the magnetic vector potential normal to the boundary is continuous and $\vec{A}^{\perp}_{\rm above} = \vec{A}^{\perp}_{\rm below}$.