The lemma is as follows
Let $f:X\to N$ be a smooth mapping, where $X$ is $m$ dimensional and $N$ is $n$ dimensional. Moreover, $m\geq n$. If $y\in N$ is a regular value, both for $f$ and the restriction $f|\partial X$, then $f^{-1}(y)\subset X$ is a smooth $(m-n)$ manifold with boundary. Furthermore the boundary $\partial(f^{-1}(y))$ is precisely equal to the intersection of $f^{-1}(y)$ with $\partial X$.
1) In the proof of the lemma he says 'Since we have to prove a local property, it suffices to consider the special case of a map $f:\mathbb H^m\rightarrow \mathbb R^n$'. I don't see how this is true. Also what exactly is meant by local property?
2)'$f^{-1}(y)$ is a smooth manifold in the neighbourhood of $x$'. What does it mean to be smooth manifold in the "neighbourhood" of $x$ ?
3)A point $x$ was taken from boundary. Another smooth map $g:U \to \mathbb{R}^n$ was constructed on a neighborhood $U$ of ${x}$ which agrees with $f$ on $U \cap H^m$. Then it was stated that we can replace $U$ by a sufficiently small neighborhood around $\bar{x}$ on which $g$ will have no critical point. Hence $g^{-1}(y)$ will be a smooth $(m-n)$-manifold. Why do we require a neighbourhood on which there is no critical point?
1) Smoothness is a local property, that's why you can always restrict to a chart (or local coordinates). Moreover, if you're not yet talking about the boundary then everything else is an open set i.e it can be mapped to the upper half space. Here that's $\mathbb{H}^m$.
2) Think of a point x in the plane, let's say the center of an open disk. Now take a smaller disk in this region, say $B_y(x) = \{y: |x-y|<\epsilon\}$. This is a smooth manifold in this neighborhood (the disk) of the point x.
3) If $y$ is a regular value, then by definition each point $p \in f^{-1}(y)$ is regular i.e $J_f(p)$ has full rank which implies no point is critical.