Let $X$ be a variety over a field $k = \bar{k}$. Assume that the Picard scheme $\text{Pic}_{X/k}$ exists and represents the functor Pic on the étale topology. Then we have the tangent space $T_0 \text{Pic}_{X/k}$ is isomorphic to $H^1(X,\mathcal{O}_X)$. My reference for this is Theorem 9.5.11 of FGA Explained by Fantechi et al.
My question relates to a step in the proof of this. Namely, let $X_{\epsilon}$ be the base change over $k$ with $X$ and $k[\epsilon]/(\epsilon^2)$. We have an exact sequence of sheaves of abelian groups:
$$0\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}^*_{X_{\epsilon}}\rightarrow \mathcal{O}_X^{*}\rightarrow 1,$$
where the first map sends $b\rightarrow 1+b\epsilon$. This is a split-exact sequence: we can send an element $b$ of $\mathcal{O}_X^{*}$ to $b+0\epsilon$.
It is then claimed that one can take cohomology to get a split exact sequence
$$0\rightarrow H^1(\mathcal{O}_X)\rightarrow H^1(\mathcal{O}^*_{X_{\epsilon}}) \rightarrow H^1(\mathcal{O}_X^{*})\rightarrow 1.$$
Why is this the case? For example, the long exact sequence of cohomology combined with this would then imply that $H^0(\mathcal{O}_X^{*}) = \text{ker}(H^1(\mathcal{O}_X)\rightarrow H^1(\mathcal{O}_{X_{\epsilon}})) = 0$, no? But this cannot be right, as we should have $k^*$ contained in the invertible global sections.
Is this a general feature of split exact sequences of sheaves and cohomology?
Cohomology (in any degree) is an additive functor: it takes direct sums to direct sums. Since a split short exact sequence is just a disguised direct sum, $H^i$ preserves split short exact sequences.
You seem to be confused about what happens on $H^0$. Just as for $H^1$ (or for any $H^i$) we get a short exact sequence of $H^0$'s, which in this case is
$$0 \to k \to k[\epsilon]^* \to k^* \to 0.$$
The boundary maps in the long exact sequence are all zero.