Theorem in question is:
Let $\mathfrak o$ be a complete local ring with maximal ideal $\mathfrak m$. Let $$ f(X) = \sum_{i = 0} a_iX^i $$be a power series in $\mathfrak o [[X]]$ such that not all $a_i$ lie in $\mathfrak m$. Let $a_0, \ldots, a_{n-1} \in \mathfrak m$ and $a_n \in \mathfrak o^\times$ is a unit. Given $g \in \mathfrak o [[X]]$ we can solve the equation $$g = qf + r$$ uniquely with $q \in \mathfrak o [[X]], r \in \mathfrak o [X], \deg r < n$.
For those who have access to book, this is chapter IV (Polynomials), subchapter 9 (Power series). For those without a book, I'll transcribe Lang's proof:
Let $\alpha$ and $\tau$ be projections on the beginning and tail end of the series, given by: $$ \alpha : \sum b_i X^i \to b_0 + b_1 X + \ldots b_{n-1}X^{n-1},$$ $$ \tau : \sum b_i X^i \to b_n + b_{n+1} X + b_{n+2}X^2 + \ldots .$$
Notice:
- $\tau(X^n h) = h$ for all $h \in \mathfrak o[[X]]$.
- $h$ is a polynomial of $< n$ if and only if $\tau(h) = 0$.
The existence of $q, r$ is equivalent with a condition that there exists $q$ such that $$ \tau(g) = \tau(qf).$$
We can write $f = \alpha(f) + X^n \tau(f)$, hence our problem is equivalent with solving $$ \tau(g) = \tau(qf) = \tau(q\alpha(f)) + \tau(q X^n \tau(f)) = \tau(q \alpha(f)) + q\tau(f),$$ by using (1) in the last equality.
Notice that $\tau(f)$ is invertible. Writing $Z = q\tau(f)$, the equation above is equivalent with:
$$ \tau(g) = \tau \left(Z \frac{\alpha(f)}{\tau(f)}\right) + Z = \left(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\right)Z$$
Here comes my issue with this proof.
Note that $\alpha(f) \in \mathfrak m[[X]]$, so that image of $\tau \circ \frac{\alpha(f)}{\tau(f)}$ is in $\mathfrak m[[X]]$.
We can therefore invert to find $Z$, namely $$ Z = \left(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\right)^{-1}\tau(g).$$
It is not at all clear to me why this remark about image of $\tau \circ \frac{\alpha(f)}{\tau(f)}$ implies invertibility of the function $\bigl(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\bigr)$. Could we write the inverse in a more concrete way?
Yes, you can write it concretely: if $T$ is a linear transformation, we should have $$(I+T)^{-1} = I - T + T^2 - T^3 + ...$$ where the right side makes sense, i.e. where given $Z$ the series $$Z - TZ + T^2Z - T^3Z + ...$$ converges. In this context, the observation about $T = \frac{\alpha(f)}{\tau(f)} \tau $ is that $T^n Z \in \mathfrak{m}^n[[X]]$, so that the series $Z - TZ + T^2Z - ...$ converges $\mathfrak{m}$-adically in the complete ring $\mathfrak{o}$.