The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure.
We will derive them in the case of large $n$ since this is when the Porter-Thomas distribution takes the exponential form given in the question. Also, this case admits an intuitive proof backed by a geometric picture. For small $n$, Porter-Thomas distribution is a Beta distribution. In this case, the proof turns into a lengthy calculation.
Consider a Haar-random quantum state $|\psi\rangle$ of $n$ qubits. Let $N=2^n$. Commonly, $|\psi\rangle$ is thought of as a complex vector $(a_0+ib_0, a_1+ib_1, \dots, a_{N-1}+ib_{N-1})^T \in \mathbb{C}^{N}$ of unit norm, but we will think of it as a real vector
$$
v = \sqrt{2N}(a_0, b_0, a_1, b_1, \dots, a_{N-1}, b_{N-1})^T\in\mathbb{R}^{2N}.
$$
Since $|\psi\rangle$ is drawn from the Haar measure, $v$ is uniformly distributed over a sphere of radius $\sqrt{2N}$ in $\mathbb{R}^{2N}$. We would like to characterize the distribution of the real coordinates $a_j$ and $b_j$.
An easy observation is that all coordinates $a_j$ and $b_j$ have the same distribution. This follows from the fact that Haar measure is unitarily invariant and permutations are unitary matrices. We can say more using the following
Theorem (Diaconis-Freedman). The first $k=o(d)$ coordinates of a point uniformly distributed over the surface of the $d$-sphere of radius $\sqrt{d}$ are independent standard normal variables in the limit of $d\to\infty$.[1]
In our case, permutation invariance means that any $k=o(N)$ coordinates of $v$ are independent standard normal variables, $a_j\sqrt{2N} \sim \mathcal{N}(0, 1)$ and $b_j\sqrt{2N} \sim \mathcal{N}(0, 1)$. Consequently,
$$
2Np_j = 2N|\langle j|\psi\rangle|^2 = \left(a_j\sqrt{2N}\right)^2 + \left(b_j\sqrt{2N}\right)^2
$$
is the sum of squares of two standard normal variables. In other words, $2Np_j$ is a chi-square random variable with two degrees of freedom which is the same distribution as the exponential distribution with rate parameter $\lambda = \frac{1}{2}$. Thus, the probability density function of $2Np_j$ is $\frac{1}{2}\exp(-\frac{1}{2}p)$, so probability density function of $p_j$ is $N\exp(-Np)$, establishing fact 1 in the question.
Now suppose we independently draw two quantum states $|\psi_1\rangle$ and $|\psi_2\rangle$ from the Haar measure. Fix an output bitstring $j \in \{0, 1\}^n$ and consider the output probabilities $|\langle j|\psi_1\rangle|^2$ and $|\langle j|\psi_2\rangle|^2$. Since $|\psi_1\rangle$ and $|\psi_2\rangle$ have been drawn independently from the Haar measure, the two probabilities are independent from each other. Moreover, by the arguments above both have the same distribution with density function $N\exp(-Np)$, establishing fact 2 in the question.
The key point to explain the symmetry between facts 1 and 2 is the independence of different coordinates of a uniformly distributed point on a sphere. As long as we only have access to a small number $k=o(2^n)$ of output probabilities of a Haar-random quantum state they are all independent and identically distributed random variables with Porter-Thomas distribution. In other words, for any quantum states $|\psi_j\rangle$ chosen independently from the Haar measure and any $k=o(2^n)$ bitstrings $z_i\in\{0, 1\}^n$, also chosen independently, each of the probabilities $|\langle z_i|\psi_j\rangle|^2$ constitutes an independent sample from the same distribution with density function $N\exp(-Np)$. This highlights the very high degree of symmetry of the Haar measure.
The observation that some states with non-zero probability density, such as $|0\rangle^{\otimes n}$, do not exhibit Porter-Thomas output distribution is correct. One might be tempted to dismiss this case as a zero-measure set. However, there is a small, but positive measure set of states in the vicinity of $|0\rangle^{\otimes n}$ that also do not exhibit Porter-Thomas output distribution.
The key point is that typical Haar-random states do. This can probably be made more rigorous by deriving an appropriate concentration inequality, e.g. bounding the probability that the total variation distance between the output distribution of a Haar-random state and the Porter-Thomas distribution exceeds a threshold. However, the following informal argument illustrates the point. One way to think of the process of drawing a Haar-random state is as a long sequence of draws of the real and imaginary parts of each amplitude from (approximate) standard normal distribution: $a_0, b_0, a_1, b_1, \dots$. A quantum state in the vicinity of $|0\rangle^{\otimes n}$ can be thought of as a sequence in which $a_0$ is approximately $1$ and all other numbers are approximately $0$. Probability of obtaining such a result from a sequence of draws from (approximate) standard normal distribution is extremely small, because the dimension of the Hilbert space and thus the length of the sequence is very large.
[1] Persi Diaconis and David Freedman. A dozen de Finetti-style results in search of a theory. Ann. Inst. H. Poincar´e Probab. Statist., 23(2, suppl.), p.397–423, 1987.
