How to Kill Homotopy Groups Using Framed Cobordism?

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:

1. X is framed cobordant to $N$ in $M$.
2. The induced map $\pi_2(X)\rightarrow \pi_2(M)$ is the $0$-map.
3. $\pi_1(X)$ 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.$\square$.