(1.36) $F(0)=[\delta_0]$ and $\delta_0(t)=e^0=1$ is the constant loop, so $F$ takes the identity to the identity.

(2.23) To see that $F(m+n)=F(m)\cdot F(n)$, we need to check that $\delta_n\cdot\delta_m\simeq\delta_{m+n}$ [NOTE: I got my concatenation the wrong way around in the video]. Let $p\colon\mathbf{R}\to S^1$ be the covering map $p(x)=e^{ix}$ and let $\tilde{\delta}_n$ be the unique lift of $\delta_n$ with the initial condition $\tilde{\delta}_n(0)=0$, in other words $\delta_n(t)=e^{i\tilde{\delta}_n(t)}$.

(4.03) Let $\tilde{\delta}_m$ be the unique lift of $\delta_m$ to the same covering space with $\tilde{\delta}_m(0)=\tilde{\delta}_n(1)$. The concatenation $\tilde{\delta}_n\cdot\tilde{\delta}_m$ makes sense and is a lift of $\delta_n\cdot\delta_m$.

(5.23) Finally, let $\tilde{\delta}_{m+n}$ be the unique lift of $\delta_{m+n}$ to the same covering space with $\tilde{\delta}_{m+n}(0)=0$.

(6.28) We have

$$\begin{align*} \tilde{\delta}_{m+n}(t)&=2\pi(m+n)t)\\ \tilde{\delta}_m(t)&=2\pi mt\\ \tilde{\delta}_n(t)&=2\pi(nt+m) \end{align*}$$ so $\tilde{\delta}_n\cdot\tilde{\delta}_m$ and $\tilde{\delta}_{m+n}$ are paths which start at $0$ and end at $2\pi(m+n)$.

(9.16) Since $\mathbf{R}$ is simply-connected, any two paths with the same endpoints are homotopic rel endpoints, so $\tilde{\delta}_m\cdot\tilde{\delta}_n\simeq\tilde{\delta}_{m+n}$ via a homotopy $H$. The homotopy $p\circ H$ is then a homotopy $\delta_n\cdot\delta_m\simeq\delta_{m+n}$.

(10.30) As $F$ is a homomorphism, it suffices to show that $n\in\ker(F)$ implies $n=0$. We therefore want to show that if $n\neq 0$ then $\delta_n$ is not homotopic to the constant loop.

(11.18) For this, we will show that, for a suitable covering space, the monodromy $\sigma_{\delta_n}$ is nontrivial. Let $p\colon\mathbf{R}\to S^1$ be the covering space $p(x)=e^{ix}$. For each $2\pi k\in 2\pi \mathbf{Z}=p^{-1}(1)$, the path $\tilde{\delta}_n(t)=2\pi(nt+k)$ is a lift of the loop $\delta_n$ with initial condition $2\pi k$, so

$$\sigma_{\delta_n}(2\pi k)=\tilde{\delta}_n(1)=2\pi(n+k).$$ We see that the monodromy $\sigma_{\delta_n}\colon 2\pi\mathbf{Z}\to 2\pi\mathbf{Z}$ is $2\pi k\mapsto 2\pi(k+n)$, which is nontrivial if $n\neq 0$.

(15.07) Given a loop $\gamma\colon[0,1]\to S^1$ based at $1\in S^1$, we want to find an $n$ such that $\delta_n\simeq\gamma$. Take the same covering space $p\colon\mathbf{R}\to S^1$ and lift $\gamma$ to get a path $\tilde{\gamma}$ in $\mathbf{R}$ with $\tilde{\gamma}(0)=0$ (so $\gamma(t)=\exp(i\tilde{\gamma}(t))$). If $n=\tilde{\gamma}(1)/2\pi$ then $n\in\mathbf{Z}$. Let $\tilde{\delta}_n$ be the unique lift of $\delta_n$ with $\tilde{\delta}_n(0)=0$. Then $\tilde{\delta}_n$ and $\tilde{\gamma}$ are paths in $\mathbf{R}$ connecting $0$ to $2\pi n$. Because $\mathbf{R}$ is simply-connected, these paths are homotopic via a homotopy $H$, which implies that $\delta_n\simeq\gamma$ via the homotopy $p\circ H$.