How to Kill Homotopy Groups Using Framed Cobordism? Let $M$ be an orientable manifold (with or without boundary), $N$ a framed submanifold in the interior of $M$ and assume (if necessary) that $\dim N<(\dim M)/2$. 
1. If some low-dimensional homotopy group $\pi_i(N)$ of $N$ is nonzero as well as its image in $\pi_i(M)$,
is it possible to find a $N'$ framed cobordant to $N$ in $M$ such that $\pi_i(N)$ (or at least its image in $\pi_i(M)$) is zero? 
In other words, can I use framed cobordisms and framed surgeries in $M$ to transform $N$ to something $k$-connected?
2. Also, could you recommend a book which could give me more intuition on this kind of questions? 
I just have read Milnor's Topology from Differentiable Viewpoint..
 A: The answer to "can I use framed cobordisms and framed surgeries in M to transform N to something k-connected" is no and I'll provide an answer below.  In fact, what I'll show is the following:

There are $N$ and $M$ as in the problem statement, with $\pi_2(N)\rightarrow \pi_2(M)$ non-trivial, but for which there is no closed manifold $X$ meeting all of the following requirements:

*

*X is framed cobordant to $N$ in $M$.

*The induced map $\pi_2(X)\rightarrow \pi_2(M)$ is the $0$-map.

*$\pi_1(X)$ is finite.


I don't see how to get rid of condition 3, but there may be a way.  Getting rid of it would answer your first question negatively.  Also, no where in my proof do I use the fact that $X$ is framed cobordant to $N$ in $M$.  I use that it is framed, and that it is cobordant to $N$.
Consider $N:=\mathbb{C}P^2 \times \{p\}\subseteq M:=\mathbb{C}P^2\times S^n$ for $n > 4$.  Due to the product structure, $N$ is framed, and further, the inclusion $i:N\rightarrow M$ induces an isomorphism on $\pi_2$.  I'll show there is no closed $4$-manifold meeting all three of the above requirements.
Suppose for a contradiction that such an $X$, with inclusion map $j$, exists.
Lemma 1:  The space $X$ must be orientable with non-trivial first Pontrjagin class.
Proof:  We have $j^\ast(TM) = TX\oplus \nu$ where the normal bundle $\nu$ is trivial since $X$ is framed.  Then $w_1(TX) = w_1(j^\ast(TM)) = j^\ast(w_1(TM)) = 0$, so $X$ is orientable.
Now, as $X$ is cobordant to $\mathbb{C}P^2$, the Pontrjagin numbers must agree. Since $p_1(\mathbb{C}P^2) \neq 0$, it now follows that the same is true of $X$.  $\square$
Let $\pi:\overline{X}\rightarrow X$ denote the universal cover of $X$.
Lemma 2:  The first Pontrjagin class of $\overline{X}$ is non-trivial.
Proof:  The projection $\pi:\overline{X}\rightarrow X$ induces an isomorphism $\pi^\ast(TX)\cong T\overline{X}$.  Since $X$ is orientable and has finite fundamental group (say, of order $s$), it follows that the induced map $\mathbb{Z}\cong H^4(X)\rightarrow H^4(\overline{x})\cong \mathbb{Z}$ is multiplication by $s$.  Thus, $p_1(T\overline{X}) = \pi^\ast( p_1(TX)) = sp_1(X)\neq 0$.  $\square$
Lemma 2 above is clearly incompatible with the following lemma, which gives the contradiction to the existence of $X$.
Lemma 3:  The manifold $\overline{X}$ is stably parallelizable, so, in particular, the first Pontrjagin class of $\overline{X}$ is trivial.
Proof:  The map $j\circ \pi$ is trivial on $\pi_2$ since $j$ is trivial on $\pi_2$ by assumption.
Since the maps $\pi_2(\overline{X})\rightarrow H_2(\overline{X})$ and $\pi_2(\mathbb{C}P^2)\rightarrow H_2(\mathbb{C}P^2)$ are isomorphisms (from the Hurewicz theorem), we deduce that the map $H_2(\overline{X})\rightarrow H_2(M)$ is trivial.  Dualizing, we see the map on $H^2$ is also trivial.
It follows from my answer to this MSE question that the map $j\circ \pi$ lifts to a map $f:\overline{X}\rightarrow S^5\times S^n$.  This lifted map is homotopically trivial by cellular approximation, so it follows that $j\circ \pi$ is homotopically trivial.
Pulling back a bundle along a homotopically trivial map gives a trivial bundle, so we see $$(j\circ \pi)^\ast (TM) = \pi^\ast(j^\ast(TM)) = \pi^\ast(TX \oplus \nu) = T\overline{X}\oplus \pi^\ast(\nu)$$ is trivial.  On the other hand, the normal bundle $\nu$ is trivial since $X$ is framed and pulling back a trivial bundle gives a trivial bundle.$\square$.
