A smooth map $F: M \to N$ between two manifolds is said to be an immersion if the pushforward $F_{*} : T_{p} M \rightarrow T_{F(p)} N$ is injective for all $p \in M$. If $M$ and $F(M) \subset N$ are also homeomorphic via $F$, then $F$ is said to be a smooth embedding.
For the embedding part of my question, I think that asking for $F$ to be a homeomorphism means that we can just put the subspace topology of $N$ on the image $F(M)$ and $M$ and $F(M)$ will still be homeomorphic (I think I can see why this is a desirable thing to have, but is there anything else going on here?), that is, besides the fact that $M$ can be immersed in $N$ via $F$, $M$ and $N$ are also topologically the same.
Now, as for the injectivity of the pushforward: what's happening here? I understand that the pushforward is sort of the best linear approximation at each point of my original $F$ (locally, of course), so that injectivity of the pushforward would mean I deform my original manifold $M$ (via $F$) in a way that changes near a point $p$ occur in a unique manner (that is, I send different tangent vectors at $p$ to different tangent vectors at $F(p)$). But why is it a bad thing if $F_*$ is not injective? What does that imply, what good things are being preserved because of injectivity? The most simple example I think of are things like $\alpha(t) = (t^2, t^3)$ which fail to be immersions at $0$, but the only thing going on there is that the curve is not differentiable at $0$ (thefore we can't talk about tangent lines), but how does that generalize to other, more complex examples?
An immersion doesn't need to be injective. The nodal cubic $F:\mathbb{R}\to\mathbb{R}^2$ given by $t\mapsto (t^2-1,t^3-t)$ is a nice example of that. In fact, $F(\mathbb{R})$ is not even a manifold! With a bit more thought, you can actually prove from first principles that $F(\mathbb{R})$ is not homeomorphic to $\mathbb{R}$. However, the injectivity of the tangent space condition that defines immersions guarantees that locally $F$ is a "homeomorphism onto its image". so if you pick a small interval $(a,b)\subset\mathbb{R}$, then you will have $(a,b)\cong F( (a,b) )$.