The answer to the first question is yes. If A and B have a direct sum A ⊕ B in C, then there are inclusions iA : A → A ⊕ B, iB : B → A ⊕ B and projections pA : A ⊕ B → A, pB : A ⊕ B → B such that pAiA = 1, pBiB = 1, and iApA + iBpB = 1. Conversely, the existence of such maps in an Ab-enriched category make A ⊕ B a direct sum of A and B, even if we do not asssume a priori that A and B have a direct sum. Now if we form the quotient C/I by an ideal I, and two objects A and B with a direct sum A ⊕ B in C, the image of this system of maps presents the image of A ⊕ B as the direct sum of the images of A and B. In short, direct sums are absolute colimits, and as the quotient functor C → C/I is essentially surjective (indeed, bijective on objects), every pair of objects of C/I inherits a direct sum from C.
Since $A$ is an abelian category, it is in particular additive, i.e. has a zero object, a product functor: $A\times A\rightarrow A$, and is Ab-enriched, i.e. hom sets are abelian groups. Since $B$ is a full subcategory, it is also Ab-enriched. If $B$ is closed under finite direct sums, then $B$ also contains the zero object (via an empty direct sum). According to page 5 of Weibel's An Introduction to Homological Algebra, finite direct sums are the same as finite products in $A$. Hence, if $B$ is closed under finite direct sums then it also has a product functor: $B\times B\rightarrow B$. Thus, before we even place (1)-(3) we know $B$ is additive.
Now, if $B$ satisfies (3) then it's an abelian subcategory because it's abelian and exact sequences in $B$ are still exact in $A$ (see page 7 of Weibel). Of course, it's enough to verify that short exact sequences in $B$ are still exact in $A$, and this comes down to $B$ being closed under kernels and cokernels.
A category is exact if it possesses a class of short exact sequences. See wikipedia's article for details. Every abelian category is exact and is closed under extensions. Thus, if $B$ satisfies (3) then it must also satisfy (2), under your initial assumptions on $B$.
A final note of caution: when someone says thick subcategory they usually mean (in my experience) that the ambient category is triangulated, which $A$ need not be. According to nLab, some people use the term thick subcategory if the ambient category is abelian, but there seems to be disagreement about the meaning of the term when so applied. The majority seem to use it to mean that the subcategory is closed under extensions (but not necessarily under kernels and cokernels). So...this paragraph is answering a question of terminology which you didn't ask, but which Fernando Muro's second comment brought up.
Best Answer
It is not true in general that the abelian subcategory (by which I mean sub-abelian category) generated by an object $X$ is all subquotients of finite sums of $X$. It is contained in these subquotients, but it might not be all of them. This is because, for instance, not every subobject of $X$ is the kernel of an endomorphism of $X$ (or more generally, a map from $X$ to a sum of copies of $X$).
As an example, consider the abelian subcategory of $\mathbb{Z}$-modules generated by $\mathbb{Q}$. Because any $\mathbb{Z}$-homomorphism of $\mathbb{Q}$-vector spaces is automatically $\mathbb{Q}$-linear, you get only the finite dimensional $\mathbb{Q}$-vector spaces, and don't, for instance, get the subgroup $\mathbb{Z}$ of $\mathbb{Q}$.