A particular confusion in showing $M_m$ is naturally isomorphic with $(F_m/F_m^2)^*$ I am working through the proof of a lemma in Frank Warner's Foundations of Differentiable Manifolds. I am trying to show that the maps he defines in the proof of this lemma are in fact inverses. 
Here $F_m^k$ is defined to be the set of k-fold products of germs at m, $M_m$  is the set of tangent vectors at $m$.

Attempt: Let $w\in M_m$, this is mapped to the functional in $(F_m/F_m^2)^*$ that sends $f$ to $m(f)=m^i\frac{\partial f}{\partial x^i}$. We want to show that $v_w(f)=w$.
$$v_w(f)=l(\{w(f)-w(f)(m)\})=l(\{w(f)\})-l(\{w(f)(m)\})$$
But I am not sure how to continue.
 A: A heuristic picture to keep in mind
It's a bit more natural to think about the isomorphism $T^*_mM\simeq F_m/F_m^2$. Using a local coordinate system this says that a covector $df$ is "the same" as a function $f(x)=df\cdot x$ what is "the same" as having a Taylor expansion $f(x)=f(0) + df\cdot x + O(x^2)$ for $f$ such that $f(0)=0$ (translates to "take $F_m$") and dropping $O(x^2)$ term (translates to "divide by $F_m^2$").
Lenghty explanation
Let's start with a vector $v\in T_mM$, i. e. an $\mathbb R$-linear map from $C^\infty(M)\to \mathbb R$ satisfying:
$$v(fg) = v(f)\cdot g(m) + f(m)\cdot v(g)$$
Lemma: If $f|_O=g|_O$ for some open neighborhood $O\ni m$, then $v(f)=v(g)$.
Proof: Using linearity this translates to $v(h)=0$ if $h|_O=0$. Using a bump function we can construct a smooth function $\phi$ such that $\phi(m)=1$ and $\mathrm{supp}~\phi \subseteq O$. Then the smooth function $h\cdot \phi=0$ (i.e. the zero function) on the whole of $M$. Hence,
$$0 = v(0) = v(h)\cdot \phi(m) + h(m) \cdot v(\phi) = v(h)$$
The above lemma basically says that if $(f, U)$ and $(g, V)$ represent the same germ $[f, U] = [g, V] \in C^\infty_m(M)$ (i. e. $f|_O=g|_O$ for some open $O\subseteq U\cap V$ containing $m$), then $v(f)=v(g)$. Hence, it makes sense to speak about the value $v([f, U]) := v(f)$.
It's just one step further from observing that we can define the value of a germ at point $m$ by:
$$ [f, U](m) := f(m)$$
Making the stalk $C^\infty_m(M)$ (the set of germs) into an algebra via operation on representives we discover that $v$ is a linear map and moreover it's a derivation:
$$v( [f, U]\cdot [g, V] ) = v([f, U])\cdot {[g, V]}(m) + v([g, V])\cdot {[f, U]}(m)$$
So far so good. Now define $F_m\subseteq C^\infty_m(M)$ to be the set of germs vanishing at $m$, i.e. ${[f, U]}\in F_m$ only if $f(m)=0$. $F_m$ has a structure of a vector space. It also makes sense to define $F_m^2$ as the set of sums of the form:
$${[f_1, U_1]}\cdot {[g_1, V_1]} + \dots + {[f_n, U_n]}\cdot {[g_n, V_n]}$$
where $f_i(m)=g_i(m)=0$ represent the elements of $F_m$. This also happens to be a real vector space. In fact this is a subspace of $F_m$ as such expressions vanish at $m$. So it makes sense to speak about the coset space $F_m/F_m^2$ and its dual.
Observe that if $f(m)=g(m)=0$, then $v(fg) = v(f) \cdot g(m) + v(g)\cdot f(m)=0$, what translates to the statement that $v$ is the zero map on the $F_m^2$ (this is a just space of (germs corresponding to) sums of such products of functions).
We know that $v$ is a linear map on $F_m$ (as a linear function on $C^\infty_m(M)$) and that $v(F_m^2)=0$. Hence, it induces a linear map on the quotient space:
$$\nu_v\colon F_m/F_m^2\to \mathbb R$$
given by $\nu_v([f, U]) = v(f)$.
Note that the map $v\mapsto \nu_v$ is linear.
Take any linear functional $\nu\in (F_m/F_m^2)^*$. We need to produce a vector from it -- something that takes a function living on $M$ and returns a real number. So take a function $f\in C^\infty(M)$. Subtract $f(m)$ giving us a function $\tilde f\in C^\infty(M)$. Pass to the germ ${[\tilde f, M]}\in C^\infty_k(M)$ that happens to be 0 at $M$, i. e ${[\tilde f, M]}\in F_m$. We can use the projection $F_m\to F_m/F_m^2$ to map this into a coset ${[\tilde f, M]}\in F_m/F_m^2$. And then we can apply our functional $\nu$ to produce a real number. 
To improve readability (and reduce mathematical precision) we will write:
$$v_\nu(f) = \nu([f-f(m)])$$
It's not hard to prove that $v_\nu$ is a linear map (using linearity of the operations described above) and the proof that this is a derivation is given in the posted question. This means that we can get a vector $v_\nu$ out of the linear functional $\nu$.
Observe also that the map $\nu\mapsto v_\nu$ is linear. 
We have related the tangent space $T_mM$ and the space of functionals $(F_m/F_m^2)^*$ using two linear maps. Modulo clutter (what makes things precise but less readable) we see that their compositions evaluate to identites:
$$v\mapsto \nu_v \mapsto v_{\nu_v} = v$$
$$\nu\mapsto v_\nu \mapsto \nu_{v_\nu} = \nu$$
what ends this logical detour.
A short excursion into the realm of commutative algebra 
(This is just a sketch. For better explanation see section 2.1 in the excellent lecture notes The Rising Sea of Ravi Vakil).
$F_m$ happens to be an ideal in the commutative ring $C^\infty_m(M)$. (Basically if $f(m)=g(m)=0$, then $(f+g)(m)=0$ and for every $h$ we have $(h\cdot f)(m)=0$). So we know what $F_m^2$ is and that $F_m/F_m^2$ is a module over $C^\infty_m(M)/F_m$. By writing down the exact sequence:
$$0\rightarrow F_m\rightarrow C^\infty_m(M) \rightarrow \mathbb R\rightarrow 0 $$
one can conclude $C^\infty_m(M)/F_m\simeq \mathbb R$, i. e. $F_m/F_m^2$ is a real vector space.
