The Lévy measure $\nu$ can be seen as an intensity measure of the jumps of the Lévy process we need to construct. The condition $$\int (1\land x^2)\nu(dx)$$ tells you two things:
The value of $\nu(\{|x|\geq 1\})=\int_{|x|\geq 1} \nu(dx)$ is finite. For the jumps of the Lévy process, this simply means that the number of jumps with jump size at least one is almost-surely finite. We obviously need this condition, since if this number were infinite, the sum of those jumps cannot converge. In your setting, this means $\nu_0$ is finite.
Often times the Lévy measure $\nu$ "blows up" around $0$, meaning $\nu(\{|x|<1\})=\infty$. This means there can be infinitely many "small jumps" with jump size at most $1$. This is where it gets tricky: We want the sum of the jumps $\sum_{k=0}^\infty X_k$ to converge. As you mentioned, Kolmogorovs Three Series Theorem states that it is necessary that $$\sum_{k=1}^\infty \text{Var}(X_k)<\infty$$ where $X_k$ denotes the sum of the jumps with jump size of magnitude between $1/k$ and $1/(k+1)$. As an exercise you might want to show that the amount of such jumps with size between $1/k$ and $1/(k+1)$ is indeed finite, so there is no issue of viewing $X_k$ that way.
If i am not mistaken, you are right in saying that $\text{Var}(X_k)\neq \int x^2\nu_k(dx)$. However, we have $$\text{Var}(X_k)=\mathbb E[X_k^2]-\alpha_k^2 \leq \mathbb E[X_k^2]=\int x^2\nu_k(dx).$$ As a consequence, the convergence of the sum of the $\int x^2 \nu_k(dx)$ is sufficient to show convergence of the sum of $\text{Var}(X_k)$.
Indeed, the choice of $1$ in the condition $\int(1\land x^2)\nu(dx)$ is quite arbitrary. You can choose any other $\epsilon>0$. The point of this condition, however, this that you "control" the number and the magnitude of small jumps, so that their sum is finite, and hence the Lévy process is well defined.
Edit: Additional material about Point processes
Since you have asked, here is a bit more about the connection between the jumps of the Lévy process and the measure $\nu$.
A point process $\mathcal P$ is a "random measure"/"random set function", meaning for a set $A\subset \mathbb R$, $\mathcal P(A)$ is a random variable, i.e. a random ensemble of points in $A$. They are both random in terms of their location in $A$, and random in terms of the number of points in $A$. Now, a Poisson point process is a particular type of point process, where you have a measure $\mu$ on $\mathbb R$, such that the number of points of $\mathcal P$ in $A$ has a Poisson-distribution with parameter $\mu(A)$ (there are additional assumptions, but i won't list them here). In particular, this means that if $\mu(A)<\infty$, the number of points in $A$ will be almost-surely finite.
Now to the Lévy process: We consider a point process $\mathcal P$ on $\mathbb R$ with intensity measure $\nu$. The points of $\mathcal P$ are considered as the "jumps" of the Lévy process in a time interval of length $1$ (we can generalize this for general time intervals, but that's besides the point for now). As mentioned above, the constraint $\int (1\land x^2)\nu(dx)$ ensures that $\nu(\{|x|\geq 1\})<\infty$. Therefore, the number of jumps with jump size greater than 1 will have a $\text{Poi}(\nu(\{|x|\geq 1\}))$-distribution, ensuring that the amount of those will be finite.
Similarly, for any $k$, if we consider the jumps of jump-size between $1/k$ and $1/(k+1)$, the amount of such jumps is equal to
$$\mathcal P\bigg(\big\lbrace 1/(k+1)\leq |x|\leq 1/k\big\rbrace\bigg)\sim \text{Poi}(\nu([-1/k,-1/(k+1)]\cup [1/(k+1),1/k]))$$
and hence there will also be an almost-surely finite amount of such jumps. In your example, $X_k$ denotes the sum of those jumps. We know that the sum converges, since it is only a finite sum, i.e. $X_k$ is given by $$X_k=\sum_{i=1}^\tau J_i^{(k)}$$ where the $J_i^{(k)}$ are the points of the Point process $\mathcal P$ in $\lbrace 1/(k+1)\leq |x|\leq 1/k\rbrace$ and $\tau$ is a Poisson-distributed random variable.
Best Answer
Solution
I think I found the error in the convolution integral. By exploiting the linearity of the integral, I did not see that for $n=2$ I have 2 identical terms, for $n=3$ I have 3 and so on. This is key to obtain the right factorial in the denominator. Thus I could apply the Taylor expansion above to get the desired result.
$$\int x \textrm{CPoi}_{v}(dx)= \int x e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}(dx)}{n!} \\=e^{-v(\mathcal{R})}\left[\int x v^{*0}(dx)+\int x v(dx)+\int x \frac{1}{2!}(v*v)(dx)+\dots\right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+\iint(s+z)\frac{1}{2!}v(ds)v(dz)+\dots \right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+v(\mathcal{R})\int(s)v(ds)+\dots\right] \\=e^{-v(\mathcal{R})}\int x v(dx)\left[1+v(\mathcal{R})+\dots \right ]$$