Given 2 sub-bundles of a given bundle, is their sum again a subvector bundle? Take any smooth projective irreducible algebraic curve $X$, any vector bundle $E$ on it and any two sub-bundles $E_1$ and $E_2$. Over each point $x$ of $X$ we can consider the sum $F_x$ of the vector spaces on the fibers of $E_1$ and $E_2$. (in general, this will not be a direct sum)
Then:
1) is $F$ a subbundle of $E$?
2) how is related $H^0(F)$ with $H^0(E_1)$ and $H^0(E_2)$?
 A: The answer to your first question is no:
Let $X$ be an arbitrary smooth curve and $P\in X$ with ideal sheaf $\mathfrak m_P\subset \mathscr O_X$. Let $E=\mathscr O_X\oplus \mathfrak m_P$, $E_1$ the diagonal of $\mathfrak m_P\oplus\mathfrak m_P\subset E$ and $E_2=0\oplus \mathfrak m_P\subset E$. Then there is a short exact sequence,
$$
0\to E_1 \to E \to \mathscr O_X \to 0,
$$
where the map $E \to \mathscr O_X$ is $(a,b)\mapsto a-b$ showing that $E_1\subset E$ is a subbundle ($E_2$ is obviously a subbundle). However $F$ is equal to $\mathfrak m_P\oplus \mathfrak m_P$ (and it is actually equal to $E_1\oplus E_2$) and $E/F$ is a skyscraper sheaf at $P$, so $F$ is not a subbundle of $E$.
For your second question, I think there are just too many possibilities. You have to make the question more precise.
A: Another example when a sum of two subbundles is not a subbundle is the following. Let $V$ be a vector space (say 2-dimensional) and $X = P(V)\times P(V)$. Then we have two subbundles $O(-1,0) \subset V\otimes O$ and $O(0,-1) \subset V\otimes O$ on $X$ (the pullbacks of the tautological subbundles from the factors). However, it is easy to see that summing up the embeddings one gets an exact sequence
$$
0 \to O(-1,0) \oplus O(0,-1) \to V\otimes \Delta_*O(1) \to 0,
$$
where $\Delta:P(V) \to P(V)\times P(V)$ is the diagonal embedding. In particular, the sum $F$ (which is the image of the first map) although isomorphic to the direct sum is not a subbundle (since the rank of the map drops at the diagonal).
As for the cohomology, there is an exact sequence
$$
0 \to E_1 \cap E_2 \to E_1 \oplus E_2 \to F \to 0,
$$
which gives a long exact sequence of cohomology. So, to know $H^0(F)$ you need to know first $H^0(E_1\cap E_2)$ and $H^1(E_1\cap E_2)$. Note however, that if $E_1$ and $E_2$ do not intersect generically, then $E_1 \cap E_2$ is a zero rank subsheaf of a torsion free sheaf $E_1 \oplus E_2$, hence $E_1 \cap E_2 = 0$. So, in this case $F = E_1 \oplus E_2$, hence $H^0(F) = H^0(E_1) \oplus H^0(E_2)$.
