Is there a way of connecting manifolds together in a higher dimension?

This question might be very naive since I haven't taken much topology or differential geometry yet; so please make the answer more conceptual than mathematical. We can think of a two dimensional sheet as joining an infinite number of one dimensional lines along a second dimension that doesn't exist for the lines. This also works if the surface is curved, only now, the lines are one dimensional curves.

So, in a sense, spaces can be joined together along a higher dimension if the lower dimensional spaces change slowly enough as we move along the higher dimension - or else we would not have a smooth surface at the end. If what I meant is not clear, the image below might explain it better:  In a sense, what I mean by the curves changing too fast is that after combining them, the resulting surface might not be continuous or differentiable (but I suppose I can only say this if the surface can be embedded in a euclidean space).

So my question is this: given an infinite set of manifolds (reasonably smooth or continuous and not necessarily reimannian) of dimension "n", is there a way to connect them like above in a higher dimension to give us a n+1 dimensional manifold?

If so, what conditions do the manifolds need to satisfy (conceptually)? If this question is too mathematical to answer conceptually, please list a few sources or let me know what this is called so that I can read about this myself.

This is a good question and there are a couple of different ways to make it precise, all of which lead to quite interesting mathematics. One is to ignore everything happening in the "middle" and focus on the "ends"; that is, we will settle for connecting two manifolds, forget about infinitely many for now. Then what you are talking about is known as a cobordism between two manifolds $$M, N$$, meaning a manifold in one higher dimension whose boundary is $$M \sqcup N$$. If a cobordism exists between $$M, N$$ they are said to be cobordant. As a simple example, the pair of pants is a cobordism between a circle $$S^1$$ and two circles $$S^1 \sqcup S^1$$: If you slice up a pair of pants and think of it as a "movie" going forward in time you'll note that in the middle you get a wedge of two circles, which is not a manifold. This sort of thing is investigated in Morse theory, which is a whole fascinating topic in itself and about which I will say nothing more.

As a general question we can ask:

Question: When are two $$n$$-manifolds $$M, N$$ cobordant?

You might think that this is a hard and deep problem in manifold topology. Remarkably, for closed $$n$$-manifolds this question has a more-or-less complete answer due to Thom. Associated to any such $$n$$-manifold is a bunch of numbers called the Stiefel-Whitney numbers, which take values $$\bmod 2$$, and Thom showed:

Theorem (Thom): Two closed $$n$$-manifolds $$M, N$$ are cobordant iff their Stiefel-Whitney numbers are the same.

The simplest Stiefel-Whitney number to explain is called $$w_n$$, and it's given by the Euler characteristic $$\bmod 2$$. It's a nice exercise to show that this is a cobordism invariant, meaning that if $$M$$ and $$N$$ are cobordant then $$\chi(M) \equiv \chi(N) \bmod 2$$.

Corollary: A closed manifold with odd Euler characteristic is not cobordant to a closed manifold with even Euler characteristic.

For example, the real projective plane $$\mathbb{RP}^2$$ has Euler characteristic $$1$$ and so is not cobordant to the torus $$T^2$$ (Euler characteristic $$0$$) or the $$2$$-sphere $$S^2$$ (Euler characteristic $$2$$). However, the torus and the $$2$$-sphere are cobordant (exercise), and in fact every pair of closed orientable surfaces is cobordant (exercise). This is a very special case of Thom's theorem (which can be proven much more easily than what Thom did), since it turns out that for such surfaces both of the Stiefel-Whitney numbers (which are called $$w_1^2$$ and $$w_2$$) vanish.

There's a whole lot more to say here about cobordism rings and so forth but this is a good place to stop for now, I think.