I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. I'll just be giving some examples, but they are very constructive and in my opinion they illustrate exactly what goes wrong.

As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)

As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. My short answer is that under rescaling by factor $\lambda$, derivatives scale by $\lambda$ and volumes by $\lambda^{-n}$, so integrating derivatives to the $n$ won't change under rescaling. The following examples illustrate how this affects embeddings.

As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)

As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. I'll just be giving some examples, but they are very constructive and in my opinion they illustrate exactly what goes wrong.

As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)

As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.

Hope this helps!

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. I'll just be giving some examples, but they are very constructive and in my opinion they illustrate exactly what goes wrong.

As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)

As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.