Connected sum is defined as Wiki .But I think the result of connected sum is not unique. For example ,make connected sum on $S^2$ with itself . Then , the result can be $T^2$ or Klein bottle. Is it really so ?
If it is right , why it is not be declared which kind the connected sum is ?
Because my English is very poor, so I add a picture ,although it is very rough, but maybe it is useful to understand what I confuse.
The connected sum of two manifolds $M$ and $N$ (if either is orientable, orient it) is, by definition, the following operation.
Pick embeddings $\iota_M, \iota_N: D^n \to M, N$. If the manifolds are smooth, pick these to be smooth embeddings that agree with the oreintations of $M$ and $N$, when defined. Otherwise pick them to be locally flat embeddings. Choosing these embeddings provide diffeomorphisms $f_M: S^{n-1} \to \partial(M - \text{int } \iota_M(D^n))$, and similarly with $N$; we may then define $M \# N$ to be the manifold obtained by gluing the disjoint union of $M - \text{int } \iota_M(D^n)$ and $N - \text{int } \iota_N(D^n)$ by the homeomorphism/diffeomorphism of the boundary $f_N \circ f_M^{-1}$.
When you ask whether this is well-defined, you're asking whether taking two manifolds (if neither are non-orientable, both have fixed given orientations; I'm being careful about this because it matters! $\Bbb{CP}^2 \# \Bbb{CP}^2$ and $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ are different manifolds.) and doing this disc nonsense gives you a well-defined output manifold $M \# N$, independent of the choice of disc. This is not easy! Smale proved this in the smooth case in the 60s, and in the topological case it's resolved using what's called the annulus conjecture, resolved only in full generality in 1982.
What you're actually writing about in your post is not connected sum. The connected sum of two copies of $S^2$ (each with either orientation) is $S^2$ again (with some orientation). What you were doing is apparently denoted the "autosum" (see the comments), as you're deleting two discs inside one manifold $M$, as opposed to deleting discs in two different manifolds. This is always isomorphic to either $M \# S^{n-1} \times S^1$ if the two discs you embed for this process have the same orientation; and $M \# (S^{n-1} \tilde \times S^1)$ if they have different orientation. Here $S^{n-1} \tilde \times S^1 = S^{n-1} \times [0,1]/(x,0) \sim (r(x),1)$ is the twisted product of $S^{n-1}$ and $S^1$, and $r$ is the reflection across some hyperplane; for $n=2$ this gives the Klein bottle.