I don't understand the following notation:
$$V_F := V \otimes_k F$$
First of all, I know that that the product is a bilinear operation, i.e. $A \otimes A \to A$, between elements of the vector space $A$ in the algebra, but $F$ is a field, isn't it? $K$ is indeed a subfield of the bigger field $F$ with the operation restricted like in the classical example of $\mathbb R$ and $\mathbb C$. I've found a similar question and answer for vector spaces, and it explains that
$V_K$ is spanned by symbols of the form $a \otimes v$
but there it is noted that
these rules are not enough to combine every sum into an element of the form $a \otimes v$.
Therefore here, in the more complicated case of an algebra instead of a vector space, I'm even more confused...
Secondly, is there a way to reconcile the above algebraic definition with a geometric point of view (e.g. Lie algebra in differential geometry)? Where they say
The set of left-invariant vector fields $\mathbb g$ with the Lie bracket [ , ] : $g \times g \to g$ is called the Lie algebra of a Lie group $G$.
is there an equivalent definition in, let's say, noncommutative algebra?
If $V$ is a vector space over $k$, $V_F = V \otimes_k F$ is a vector space over $F$ called the extension of scalars of $V$ to $F$, with respect to a fixed choice of embedding $f : k \to F$. It can be understood explicitly as follows: if $v_1, \dots v_n$ is a basis of $V$ over $k$ ($n$ can be infinite here), then their image in the extension of scalars $v_1 \otimes 1, \dots v_n \otimes 1$ (often just written $v_1, \dots v_n$ again) remains a basis of $V_F$ over $F$. So for example
$$k^n \otimes_k F \cong F^n$$ $$M_n(k) \otimes_k F \cong M_n(F)$$ $$k[x_1, \dots x_n] \otimes_k F \cong F[x_1, \dots x_n]$$ $$\mathfrak{sl}_n(k) \otimes_k F \cong \mathfrak{sl}_n(F).$$
(So far these are all just isomorphisms of vector spaces.)
If $V$ has the structure of a $k$-algebra (commutative, associative, Lie, etc.) then $V \otimes_k F$ inherits this structure, but now as an $F$-algebra. The $k$-linear multiplication $m : V \otimes_k V \to V$ gets upgraded to an $F$-linear multiplication $m_F : V_F \otimes_F V_F \to V_F$. Again working explicitly, if $v_1, \dots v_n$ is a basis of $V$ and $m$ has structure constants
$$m(v_i) = \sum_{jk} m_i^{jk} v_j v_k$$
with respect to this basis, then the extension of scalars $m_F$ has structure constants $f(m_i^{jk})$ with respect to $v_1, \dots v_n$ regarded as a basis of $V_F$ over $F$ as above. This makes all of the isomorphisms I just wrote down above isomorphisms of $F$-algebras.
I don't understand your second question or what it has to do with your first question.