Witt theorem proof - the case when the subspace is totally isotropic Witt extension Theorem Stronger Version. I want to ask for the case that $W$ is totally isotropic. 
This is a part of the proof of Theorem 1.5.5 in W. Scharlau: Quadratic and Hermitian Forms.

We consider the surjective map $$\pi\widehat b\colon V\to V^* \to W^*$$ where $\pi$ is the canonical projection.
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  Note that $W\subset\ker (\pi\widehat b)$.
  Let $W'$ be a subspace which is mapped isomorphically onto $W^*$.
  We claim that $W\oplus W'$ is a regular subspace.
  In fact, if $x+x'\in W\oplus W'$ with $x'\ne 0$, then there is a $y\in W$ such that $$b(x+x',y)=b(x',y)=(\widehat by)x'\ne0.$$
  If $x\ne 0$, then there is a $y\in W'$ such that $b(x,y)=(\widehat by)x\ne 0$.
  By definition of $W'$ the map $$i\colon W\oplus W' \to W\oplus W^* = \mathbb H(W) \qquad (x,y)\mapsto (x,\widehat by|_W)$$ is an isometry with $i|_W=id$.
  The same argument can be applied to $\sigma W$ and gives a subspace $W''$ and an isometry $j\colon \sigma W\oplus W'' \to \mathbb H(\sigma W)$ with $j|_{\sigma W}=id$.
  By 4.4 the map $\sigma$ can be extended to $W\oplus W'$ $$W\oplus W' \overset{i}\longrightarrow W\oplus W^* \overset{\sigma\oplus\sigma^{\ast-1}}\longrightarrow \sigma W\oplus(\sigma W)^* \overset{j^{-1}}\longrightarrow \sigma W\oplus W''.$$
  Now $m(W\oplus W')=2\dim W < 3\dim W=m(W)$ and we can apply the induction hypothesis to finish the proof.'

This is outline of proof given in the book; but I have certain doubts that why $i$  is an isometry as well as $j$. Also can someone give me explicit way for using induction in above case and give full details for what are reflections in this case ?
 A: DISCLAIMER: This is not an area I am very familiar with. (Basically what I know about this is what I learned when answering a few recent questions on this site.) So take this answer with a grain of salt. (It would definitely be nice if somebody with a good knowledge of bilinear forms and Witt's theorem could look at the question.)
The things I wrote here overlap to a great extent with the stuff that I have said when discussing with the OP in chat. Since the OP still asked for more details after that, it is probable that this answer will not be satisfactory. But maybe if the stuff is collected here in the form of the answer, it might be easier for the OP to point out what specifically is the problem and which parts of the proof need more detailed explanation.

You have asked a few times how exactly is induction used in this proof. So let me start with this.
The author is doing induction on $m(W)$.
This means that the inductive step is like this: We assume that the claim is true for any subspace $W_0$ of $V$ such that $m(W_0)<m(W)$. Using this we want to show that it is true also for $W$.
In the part of proof you are asking about, the space $W\oplus W'$ has the role of $W_0$.
Since the induction hypothesis is that the claim is true for any regular subspace with smaller $m$ and $m(W\oplus W')<m(W)$, we can use the claim for the subspace $W\oplus W'$ and the map $\sigma'$.
From the induction hypothesis we get that $\sigma'$ has an extension which is composition of reflections. And since $\sigma'|_W=\sigma$, the claim for $\sigma$ follows.
I haven't seen other proofs of this claim. The author says in the book: "In this way we can avoid some technical steps in the classical proofs." So this should probably be the clever part which helps to avoid some technical details in the totally isotropic case. 

Here are more detailed explanations of some parts of the proof.

We consider the surjective map $$\pi\widehat b\colon V\to V^* \to W^*$$ where $\pi$ is the canonical projection.

Here $\widehat b \colon V\to V^*$ is determined by $$(\widehat bx)y=b(x,y),$$ i.e., $\widehat b(x)=b(x,\cdot)$.
The map $\pi$ is simply the restriction, i.e., $\pi(f)=f|_W$. The map $\pi$ is surjective.
So we have $$\pi\widehat b(x) = b(x,\cdot)|_W.$$
The map $\widehat b$ is defined in Definition 2.3. Lemma 2.5 says that $W^\bot = \ker(\pi\widehat b)$.
By Corollary 3.2 we know that $\widehat b$ is isomorphism (since $V$ is regular).
So together we get that composition of two surjective functions $\widehat b$ and $\pi$ is again surjective.

Let $W'$ be a subspace which is mapped isomorphically onto $W^*$.

According to first isomorphism theorem, there is an isomorphism between $V/\ker(\pi\widehat b)$ and $W^*$. If we choose a basis $\mathcal B$ of $V/\ker(\pi\widehat b)$ then this gives us a basis of a subspace $W'$ as needed. (To be more precise, elements of $V/\ker(\pi\widehat b)$ are equivalence classes. From each equivalence class that belongs to the basis $\mathcal B$ we choose one vector.)
Or this is basically just another (and maybe more straightforward) way of stating the same thing: You have a surjective linear map $s\colon V\to W^*$. Choose a basis $\mathcal B'$ of $W^*$. For each element $b\in \mathcal B'$ take one vector from $s^{-1}(b)$. And this gives you a basis for $W'$. 

We claim that $W\oplus W'$ is a regular subspace. 

We want to show that $(x+x')^\bot=0$ for any vector $x+x'\in W\oplus W'$. This is proved below by considering two cases. One of them is $x'\ne0$ and the other one is $x'=0$.

In fact, if $x+x'\in W\oplus W'$ with $x'\ne 0$, then there is a $y\in W$ such that $$b(x+x',y)=b(x',y)=(\widehat by)x'\ne0.$$

Since $\pi\widehat{b}|_{W'}$ is an isomorphism, it has zero kernel. So no vector $x'\in W\setminus\{0\}$ is mapped to the zero map.

If $x\ne 0$, then there is a $y\in W'$ such that $b(x,y)=(\widehat by)x\ne 0$. 

For every vector $x\in W\setminus\{0\}$ there is a function $f\in W^*$ such that $f(x)\ne0$. Now we choose $y\in W'$ which is mapped onto $f$ by the isomorphism $\pi\widehat{b}|_{W'}$.

By definition of $W'$ the map $$i\colon W\oplus W' \to W\oplus W^* = \mathbb H(W), \qquad (x,y)\mapsto (x,\widehat by|_W)$$ is an isometry with $i|_W\ne 0$.

We have $$h(x,\widehat by|_W) = (\widehat by)x=b(y,x)=b(x,y)$$ directly from the definition of $\widehat b$.
We also need to check that $i$ is injective. This follows from the fact that it is sume of identity $W\to W$ and an isomorphism $W'\to W^*$.

The same argument can be applied to $\sigma W$ and gives a subspace $W''$ and an isometry $j\colon \sigma W\oplus W'' \to \mathbb H(\sigma W)$ with $j|_{\sigma W}=id$.
  By 4.4 the map $\sigma$ can be extended to $W\oplus W'$ $$W\oplus W' \overset{i}\longrightarrow W\oplus W^* \overset{\sigma\oplus\sigma^{\ast-1}}\longrightarrow \sigma W\oplus(\sigma W)^* \overset{j^{-1}}\longrightarrow \sigma W\oplus W''.$$

Lemma 4.4 is exactly the claim that $\sigma\oplus\sigma^{\ast-1}$ is a bijective isometry whenever $\sigma$ is a linear isomorphism.

Now $m(W\oplus W')=2\dim W < 3\dim W=m(W)$ and we can apply the induction hypothesis to finish the proof.

Let us denote $$\sigma'=j^{-1}\circ (\sigma\oplus\sigma^{\ast-1})\circ i.$$ This is a map  $\sigma' \colon W\oplus W' \to V$.
We know that $W\oplus W'$ is a regular subspace and that $\sigma'$ is an isometry. And that $m(W\oplus W')<m(W)$.
Now we apply the inductive hypothesis to the subspace $W\oplus W'$ and the map $\sigma'$. We get (from the inductive hypothesis) that $\sigma'$ can be extended to a composition of reflections. Since $\sigma'|_W=\sigma$, this is also an extension of $\sigma$. 
