Let $M$, $N$ be two manifolds of dimension $n$. Fix $m>n$. Let $i:[0,1)\times M \to \mathbb{R}^m$, $j:(1,2]\times N \to \mathbb{R}^m$ be embeddings such that for each $s\in[0,1)$, $t\in(1,2]$, $i_s:M\to \mathbb{R}^m$ and $j_t:N\to \mathbb{R}^m$ are embeddings. So $i$ and $j$ are one-parameter families of embeddings. Suppose further the limit of $i_s$ and $j_t$, namely $$ i_{1^-}:M\to \mathbb{R}^m:x\mapsto \lim_{s\to 1} i_s(x) \\ j_{1^+}:N\to \mathbb{R}^m:y\mapsto \lim_{t\to 1} j_t(y) $$ is well-defined, and their images are homeomorphic, $$ \phi:i_{1^-}(M) \cong j_{1^+}(N). $$ Note that this image might not be a manifold. Now we define $M_i$ (respectively, $N_j$) to be union of $i([0,1)\times M)$ and $i_{1^-}(M)$ (respectively, $j([0,1)\times N)$ and $j_{1^+}(N)$.) Glue $M_i$ and $N_j$ using the homeomorphism $\phi$, and let $M\sqcup_{i,j}N$ denote this topological space.
Now my question is when this space $M\sqcup_{i,j}N$ will be a $n+1$-dimensional manifold, and what condition for $i,j$ must be held in order for $M\sqcup_{i,j}N$ to be a manifold, what must happen at the limit $i_{1^-}$ and $j_{1^+}$. Also, I wonder if there is a counterexample, where $M\sqcup_{i,j}N$ is not a manifold.
For example, a pair of pants, which is a cobordism between $M=S^1$ and $N=S^1\sqcup S^1$ can be constructed by taking one-parameter of embeddings $i:[0,1):M\to\mathbb{R}^3$, $j:(1,2]:M\to\mathbb{R}^3$ so that at the limit $s,t\to 1$, their images become $S^1\wedge S^1$.
I don't think I understand your glueing precisely (you are gluing points that are not in the manifolds), but I think the answer is no, even if $M=N=\{x\}$ are just single points and the limit set is a manifold! The topologist sine curve is an embedding and you can glue two together like this: