The best definition for independence I can give you, without sweating measure theory all that much, is this:
Two random variables $X$ and $Y$ (which share the same probability space, but we don't assume have the same codomain) are independent if the following holds:
For any event $A$ for $X$ and any event $B$ for $Y$, we have
$$
P(X\in A,\ Y\in B)=P(X\in A)\cdot P(Y\in B).
$$
If you want to start talking about probability densities and things of that nature for the case when you have two random vectors of different dimensions, you're going to need some more serious theory; such things can be defined fairly easily in terms of measure theory, but not as easily otherwise.
Of course, if you have an $n$-long vector and a $k$-long vector, you can think of their joint distribution as being an $(n+k)$-long vector; in this case, assuming you are talking about continuous distributions, you could probably think of it as the condition that
$$
f_{\vec{x},\vec{y}}(x_1,\ldots,x_n,y_1,\ldots,y_k)=f_{\vec {x}}(x_1,\ldots,x_n)\cdot f_{\vec{y}}(y_1,\ldots,y_k),
$$
where $f_{\vec{x},\vec{y}}$ is the joint density function and $f_{\vec{x}}$ and $f_{\vec{y}}$ are their respective marginal densities.
Actually, your second statement is false. This would only be true if $X,Y$ are jointly normal, which you do not assume. And if you're assuming joint normality then uncorrelated=independent.
Best Answer
It is not true in general: Consider, as a counterexample in $\Bbb{R}^2$, $x_i$ each chosen as a uniform variate on $(0,1)$ and $y_i$ formed using two further uniform variates $u_i$ by $$ \begin{array}{c} y_1 = (x_1 - x_2)_{\mod 1}+u_1 \\ y_2 = (x_1 + x_2)_{\mod 1}+u_2 \\ \end{array} $$ Then knowledge of either $x_i$ alone does not affect the distribution of either $y_i$, but knowledge of the whole vector $\vec{x}$ heavily affects the distribution of the $\vec{y}$; and knowledge of $\vec{y}$ does not effect the distribution of (say) $x_1$ unless you also know something about the value of $x_2$. So every pair of components are independent, but the two vectors are not.
(Obviously, the other direction does hold: If the vectors are independent, then pairs of one variate from each must be independent.)
If $X$ and $Y$ are jointly normal distributed, then pairwise independence does imply cector independence.
The second statement is the trickiest. If $X$ and $Y$ are both normally distributed, you can try for have a similar counterexample: This time, let $x_i$ each be chosen as a unit normal and take $$ \begin{array}{c} y_1 = N[(x_1 - x_2)_{\mod 1}]\\ y_2 = N[(x_1 - x_2)_{\mod 1}] \end{array} $$ where $N[\mu]$ stands for a normal distribution about mean $\mu$ and unit variance. But here, because you do have some knowledge that $x_2$ is likely near zero, knowledge of either $y_i$ does affect the distribution of $x_1$.
The counterexample can be found, however, by replacing $(x_1 - x_2)_{\mod 1}$ with something like $$ \left( e^{(x_1-x_2)^2/2} \right)_{\mod 1} $$ (and similarly for $(x_1 + x_2)_{\mod 1}$) so that the lack of knowledge of the exact value of $x_2$ again washes out any information obtained from knoing $y_1$.