Why do we derive variance-covariance matrix of coefficients instead of variance vector? I don't understand why we have to consider covariance.
Is there any problem if I just intuitively understand Var(Y) as variance vector of Y?
 A: Why do we need covariance matrix? Well, sometimes it is simply not enough to know only variances.  If you by variance vector means the diagonal of the covariance matrix, that is,
$$
\DeclareMathOperator{\V}{\mathbb{V}}
\DeclareMathOperator{\v}{\mathbb{v}}
\DeclareMathOperator{\Cov}{\mathbb{Cov}}
\v(X) = (\V X_1, \V X_2, \dotsc, \V X_n) = \text{diag}(\V X)
$$
where $\V X$ is the covariance matrix of the random vector $X$, that is,  the $n \times n$ matrix which have for $i,j$-element the covariance between the $i$th ant $j$th elements of the vector $\Cov(X_i,X_j)$. $\v(X)$ is the $n$-vector which have for element $i$ $\V X_i$, the variance of element $i$. 
One reason we need this can be seen from the variance of linear combinations of elements of $X$. That is given by
$$
\V a^T X = \V \sum_i a_i X_i = \sum_i \sum_j a_i a_j \Cov(X_i,X_j)= \\
    a^T \V(X) a
$$
Look at the 2-vector $X=(X_1, X_2)$ and $\V (X_1 - X_2) = \V X_1 -2 \Cov(X_1,X_2) + \V X_2$.
Suppose $\V X_1 =\V X_2 = 1$ then we can show that $-1 \le \Cov(X_1,X_2) \le 1$ (*).  Using that you can show that $\V(X_1-X_2)$ is somewhere between 0 and 2 only by varying the covariance in the possible range, so indeed it makes a large difference.
(*) You can show this from the fact that the correlation coefficient is between $-1$ and $1$.
