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A linear vector space $\mathbb{V}$ is set of elements $|\psi\rangle,|\phi\rangle,|\chi\rangle...$, called vectors, defined on a field $\mathbb{F}$ of scalars $(a,b,c,...)$ such that it satisfy (among other properties) that for any two elements $|\psi\rangle,|\phi\rangle \in\mathbb{V}$, there exist a definite rule of addition such that $$|\phi\rangle +|\psi\rangle=|\psi\rangle +|\phi\rangle \in \mathbb{V}$$

My question are: 1. What kind of addition rule will be non-commutative? I want an example. 2. How can the associative law of addition fail? 3. What could be an example where $|\phi\rangle +|\psi\rangle$ does not belong to $\mathbb{V}$ but $|\phi\rangle,|\psi\rangle\in\mathbb{V}$?

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    Could you be more specific about what you class as an 'addition rule'? For example we could consider a subspace of $\mathbb{R}^{3}$ (some plane) and define $+$ as the cross product of 2 vectors. Then for any 2 vectors in the plane, their 'addition' is perpendicular so not in the space. – 2017-01-05

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Note that a priori the symbol $+$ doesn't mean anything; it is just any operation that takes two inputs and produces an output. We usually use $+$ only for operations that are commutative and associative, so any $+$ that is not associative or commutative will not, intuitively, feel like an operation that deserves to be called $+$.

1-2) An easy example of an operation that is non-commutative and non-associative is subtraction of numbers. So if the symbol $+$ means $-$ (so, for example, "$5 + 2$" in this context will be the number 3), then it is not commutative: $$ 2 +1 = 1 \neq -1 = 1 + 2, $$ and it is not associative either; $$ (1 + 1) + 1 = -1 \neq 1 = 1 + (1 + 1). $$ 3) For example, you could have (stupidly) defined $V = \{-1,0,1\}$ and have $+$ be ordinary addition. Then $1 + 1 = 2 \notin V$.

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Time travel.

Draw two commutative vector sums end to end meeting at a point and they differ by the imaginary plane. It should look like a diamond/square.