In picture below, Spin(V) is Spin group. In the last paragraph , it wants to show $\pi_1(Spin^c(V))=\pi_1(S^1)$. But I almost don't know it.
First, where is the $S^1$ in ? For 'a homotopically nontrivial loop $\gamma$ in $S^1$', if the $S^1$ is $S^1$ of $SO(V)\times S^1$, how to induces a loop in $Spin^c(V)$ ? $Spin^c(V)$ is double covering of $SO(V)\times S^1$, then, there will induce two loop in $Spin^c(V)$.
Second, why $\pi_1(Spin^c(V))$ contains $\pi_1(S^1)$ ? Seemly, it is because any loop in $Spin^c(V)$ will induces loop in $S^1$ of $SO(V)\times S^1$. But $SO(V)$ is not simply connect, why the loop will not in $SO(V)$ ?
Third,how to compose 2.4.14? I don't know the mean of compose.
Forth, how to know kernel is identified with $Spin(V) $ by 2.4.13 ?
In fact, I really know nothing about the last paragraph.I very thanks for detail explain of it.
Picture below is from 75 page of Jost's Riemannian geometry and geometric analysis.
From your other question, we have a map $\mathrm{Spin}(V)\times S^1\to \mathrm{Spin}^c(V)$ with kernel $\pm(1,1)$, which means we can write $\mathrm{Spin}^c(V)\cong \mathrm{Spin}(V)\times_{\mathbb{Z}_2}S^1$ where $(a,z)\sim(-a,-z)$, or equivalently where we have $(-a,z)\sim(a,-z)$.
At this point we can speak of $\mathrm{Spin}^c(V)$ and $\mathrm{Spin}(V)\times_{\mathbb{Z}_2}S^1$ interchangeably.
Next, there is a map $\mathrm{Spin}^c(V)\to \mathrm{SO}(V)\times S^1$ given by $(a,z)\mapsto (\overline{a},z^2)$, where by $\overline{a}$ I mean the image of $a$ under the usual double covering $\mathrm{Spin}(V)\to\mathrm{SO}(V)$. The kernel of this map is $(\pm1,\pm1)$ but $(-1,-1)\sim(1,1)$ and $(1,-1)\sim(-1,1)$ so the kernel is $\mathbb{Z}_2$, and this map can be called a double covering.
By picking the particular element $1\in\mathrm{Spin}(V)$, we can map $S^1\to \mathrm{Spin}^c(V)$ via $z\mapsto (1,z)$. This is injective, i.e. $(1,z_1)\sim(1,z_2)\Rightarrow z_1=z_2$. Thus, any loop in $S^1$ (i.e. based map $\gamma:S^1\to S^1$) can be turned into a loop in $\mathrm{Spin}^c(V)$, namely the composition $S^1\xrightarrow{\gamma}S^1\to\mathrm{Spin}^c(V)$.
(Recall the composition of two functions $g:X\to Y$ and $f:Y\to Z$ is $f\circ g:X\to Z$.)
We can compose $\mathrm{Spin}^c(V)\to\mathrm{SO}(V)\times S^1$ with $\mathrm{SO}(V)\times S^1\to S^1$ (the projection map) to obtain a map $\mathrm{Spin}^c(V)\to S^1$. We may then "chase" an element through the following:
$$ \begin{array}{ccccccc} S^1 & \longrightarrow & \mathrm{Spin}^c(V) & \longrightarrow & \mathrm{SO}(V)\times S^1 & \longrightarrow & S^1 \\ z & \mapsto & (1,z) & \mapsto & (\mathrm{id},z^2) & \mapsto & z^2 \end{array} $$
Thus, if we start with a nontrivial loop $\gamma$ in $S^1$, then put it inside $\mathrm{Spin}^c(V)$, then project it back to $S^1$ again, we get $2\gamma$ which is again nontrivial. If the image of $\gamma$ inside $\mathrm{Spin}^c(V)$ were trivial then it would project onto a trivial loop in $S^1$, but it doesn't, so it is nontrivial within $\mathrm{Spin}^c(V)$.
We've seen that loops in $S^1$ induce loops in $\mathrm{Spin}^c(V)$. Any homotopy between loops in $S^1$ can be performed in $\mathrm{Spin}^c(V)$ as well, so really homotopy classes of loops in $S^1$ induce homotopy classes of loops in $\mathrm{Spin}^c(V)$, i.e. we have a map $\pi_1(S)\to\pi_1(\mathrm{Spin}^c(V))$. Since composition of loops in $S^1$ corresponds to composition of loops in $\mathrm{Spin}^c(V)$, this is in fact a group homomorphism, and we've seen its kernel is the trivial loop only, so it is an embeddding and we may identify $\pi_1(S^1)$ with a subgroup of $\pi_1(\mathrm{Spin}^c(V))$.
Suppose we had a loop in $\mathrm{Spin}^c(V)$ that didn't come from $S^1$, i.e. is not in this copy of $\pi_1(S^1)$. The text seems to be assuming that, without loss of generality, this loop's projected image in $S^1$ is trivial, but I don't see why.
At any rate, suppose we have a loop in $\mathrm{Spin}^c(V)$ which maps to the trivial loop in $S^1$. Then the loop is (up to homotopy) entirely contained within the kernel of $\mathrm{Spin}^c(V)\to S^1$, which is all $(a,z)$ with $z^2=1$, i.e. $(a,\pm1)$, but $(a,-1)\sim(-a,1)$ so kernel is exactly the set of all $(a,1)$ with $a\in\mathrm{Spin}(V)$. This is an embedded copy of $\mathrm{Spin}(V)$ within $\mathrm{Spin}^c(V)$, which is simply connected if $\dim V>2$, so this loop is trivial.