A little background, then I'll try to answer question 1.

While attempting to encode homotopy types of connected CW-complexes $K=(K^n)_{n\in\mathbb N}$ in a purely algebraic way, J.H.C Whitehead invented crossed modules (Combinatorial homotopy. II, Bull. Am. Math. Soc. 55, 453-496 (1949). ZBL0040.38801) as a component of his `homotopy systems' wherein crossed modules describe the relationship between $\pi_2(K^2,K^1,k_0)$ and $\pi_1(K^1,k_0)$. Specifically, the map $$\partial:\pi_2(K^2,K^1,k_0)\longrightarrow\pi_1(K^1,k_0)\,,$$ which takes an embedded disk $\phi:D^2\to K$ and returns the restriction $\phi|_{\partial D^2}:S^1\to K$, gives rise to a crossed module structure with respect to the conjugation action of $\pi_1(K^1,k_0)$ on $\pi_2(K^2,K^1,k_0)$.

The theorem of Mac Lane and Whitehead (On the 3-type of a complex, Proc. Natl. Acad. Sci. USA 36, 41-48 (1950). ZBL0035.39001) is that when $\pi_n(K)=0$ for $n>2$, this crossed module structure uniquely determines the homotopy type of $K$, and moreover every crossed module can be realized by such a complex.

Using this theorem, we can assume that $H=\pi_2(K^2,K^1,k_0)$ and $G=\pi_1(K^1,k_0)$ for some connected CW-complex $K$. If you aren't familiar with CW-complexes, you can think of them as a particularly nice class of topological spaces. In this setting, I can begin to explain why the braiding is not so surprising.

1. The second homotopy group of a space is normally abelian, but the relative group $\pi_2(K^2,K^1,k_0)$ is not abelian. The problem is that the boundaries are not the constant loop $S^1\to k_0$, but rather they are loops $S^1\to K^1$ based at $k_0$. The standard topological trick for showing commutativity $\alpha*\beta=\beta*\alpha$ involves sliding $\alpha$ and $\beta$ around each other. In other words, the trick involves conjugating $\alpha$ by the loop $\partial\beta$ to produce the class $\alpha^{\partial\beta}$; it's just that conjugating by the trivial loop never does anything. Thus in our group the best we can get is that $\alpha*\beta = \beta*\alpha^{\partial\beta}$, which is your equation 2.

Since this equation literally comes from passing one map in front of the other to swap their order, it's not surprising that it encodes some braiding information. That a `group with braiding' should induce a braiding on its own representations is also natural.

2. I can't quite answer this one. The representations you describe are not simply a 2-group acting on a category, at least not in any way I can make sense of. They appear to be classical representations of the algebra $H^*\rtimes G$ where $H^*=\mathsf{Fun}(H,\mathbb C)$, and where the $\rtimes$ indicates $$(\delta_m\otimes g)\cdot(\delta_n\otimes h)=(\delta_m\cdot\delta_{n^g})\otimes(gh)\,.$$

A little background, then I'll try to answer question 1.

While attempting to encode homotopy types of connected CW-complexes $K=(K^n)_{n\in\mathbb N}$ in a purely algebraic way, J.H.C Whitehead invented crossed modules (Combinatorial homotopy. II, Bull. Am. Math. Soc. 55, 453-496 (1949). ZBL0040.38801) as a component of his `homotopy systems' wherein crossed modules describe the relationship between $\pi_2(K^2,K^1,k_0)$ and $\pi_1(K^1,k_0)$. Specifically, the map $$\partial:\pi_2(K^2,K^1,k_0)\longrightarrow\pi_1(K^1,k_0)\,,$$ which takes an embedded disk $\phi:D^2\to K$ and returns the restriction $\phi|_{\partial D^2}:S^1\to K$, gives rise to a crossed module structure with respect to the conjugation action of $\pi_1(K^1,k_0)$ on $\pi_2(K^2,K^1,k_0)$.

The theorem of Mac Lane and Whitehead (On the 3-type of a complex, Proc. Natl. Acad. Sci. USA 36, 41-48 (1950). ZBL0035.39001) is that when $\pi_n(K)=0$ for $n>2$, this crossed module structure uniquely determines the homotopy type of $K$, and moreover every crossed module can be realized by such a complex.

Using this theorem, we can assume that $H=\pi_2(K^2,K^1,k_0)$ and $G=\pi_1(K^1,k_0)$ for some connected CW-complex $K$. If you aren't familiar with CW-complexes, you can think of them as a particularly nice class of topological spaces. In this setting, I can begin to explain why the braiding is not so surprising.

1. The second homotopy group of a space is normally abelian, but the relative group $\pi_2(K^2,K^1,k_0)$ is not abelian. The problem is that the boundaries are not the constant loop $S^1\to k_0$, but rather they are loops $S^1\to K^1$ based at $k_0$. The standard topological trick for showing commutativity $\alpha*\beta=\beta*\alpha$ involves sliding $\alpha$ and $\beta$ around each other. In other words, the trick involves conjugating $\alpha$ by the loop $\partial\beta$ to produce the class $\alpha^{\partial\beta}$; it's just that conjugating by the trivial loop never does anything. Thus in our group the best we can get is that $\alpha*\beta = \beta*\alpha^{\partial\beta}$, which is your equation 2.

Since this equation literally comes from passing one map in front of the other to swap their order, it's not surprising that it encodes some braiding information. [That a 'group with braiding' should induce a braiding on its own representations is also natural.] edit The previous statement was glib, so let me clarify how it works in this case. For these representations the $\delta_h$ are idempotent, so they act like projections on any given representation. This allows us to produce a canonical decomposition $V_1\cong\bigoplus_{h\in H} \delta_hV_1$, and we can think of a vector $\delta_hv_1$ as being concentrated in the $h$-component, or if you like, homogeneous of degree $h$.

Now that we have a way of decomposing a vector into components that are labelled by the elements of $H$, we can ask what happens when $v_2$ passes under each of these components individually. From the previous discussion, this should involve something like 'conjugating by $h$'. Since it doesn't make sense to conjugate a vector in a representation, this normally wouldn't work, but in our case $\partial h$ is a version of `conjugating by $h$' that lives in $G$... which also happens to act on our vector space. So we posit $$\delta_hv_1\otimes v_2\mapsto (\partial h)v_2\otimes \delta_hv_1\,.$$ I admit that this guess is not fully justified, but I hope it's at least plausible. Next we use the fact that $1=\sum_{h\in H}\delta_h$ to extend the map to inhomogeneous vectors to arrive at the formula you have given. \begin{align*} v_1\otimes v_2&=\left(\sum_{h\in H}\delta_hv_1\right)\otimes v_2\\ &=\sum_{h\in H}\delta_hv_1\otimes v_2\\ &\mapsto \sum_{h\in H}(\partial h)v_2\otimes \delta_hv_1\,. \end{align*}

2. I can't quite answer this one. The representations you describe are not simply a 2-group acting on a category, at least not in any way I can make sense of. They appear to be classical representations of the algebra $H^*\rtimes G$ where $H^*=\mathsf{Fun}(H,\mathbb C)$, and where the $\rtimes$ indicates $$(\delta_m\otimes g)\cdot(\delta_n\otimes h)=(\delta_m\cdot\delta_{n^g})\otimes(gh)\,.$$