Given a set of linearly independent vectors $\{v_1,v_2, ..., v_N\}$. (Linearly independent over the real numbers.) How can we justify that $iv_1$ is not in Span ($\{v_1,v_2, ..., v_N\}$) ? Intuitively, I would think that a linear combination of $v_1,v_2, ..., v_N$ can at best be a real number times $v_1$. Surely, if we assume that $iv_1$ is in the span there must exist unique real numbers $a_1,a_2,...,a_N$ such that $a_1 v_1+a_2 v_2+...+a_N v_N = iv_1$, so that $a_1 v_1 - iv_1 + a_2 v_2 + ... + a_N v_N = 0$ such that $\{(a_1-i),a_2,...,a_N\}$ is a set of weights that disprove that $\{v_1,v_2, ..., v_N\}$ is a vector space over the complex numbers, but $\hat{a}$ doesn't get us very far as it was the field of real numbers we were interested in.
In other words, how do we justify that a vector space over some field $K$ does not contain an element on the form $tR$, where $R$ is in $V$, but t is in some field different (but perhaps being a superset) of $K$
Is this this justified by convention, definition, or can this element also be in $V$?
:)
This is simply not true in general: consider $N=2$, $v_1=1$ and $v_2=i$ in $\mathbb{C}$.
Then the family $\left\lbrace v_1, v_2\right\rbrace$ is independent over the real numbers, yet $i v_1$ does lie in Span$\left\lbrace v_1, v_2\right\rbrace$ since $i v_1=v_2$.
Now regarding your last question:
Simply put, $tR$ does not lie in $V$ because $tR$ does not make sense until you define it. To do that, you can consider a tensor product of $V$ with a larger field, it does exactly what you want (extend the ring/field of scalars while preserving the existing structure). In that case you may want to insert a symbol $\otimes$ as in "$t\otimes R$", to make it clear that $R$ initially lies in a vector space whose coefficient field was extended into something that contains $t$.