All of my geometric intuition for "immersed" versus "embedded" surface is contained in my geometric intuition for "immersions" versus "embeddings". In particular, as many people have pointed out, immersions need not be injective. But, of course, even injective immersions need not be embeddings. As a very simple example, consider the map $f : (-\pi, \pi) \to \mathbb{R}^2$ given by
$$
f(t) = (\sin t, \sin 2t).
$$
The image of this map is a sort of "figure 8" in the plane, traced out starting at the origin, moving through quadrants II, III, I, and IV, in that order, as $t$ moves from $-\pi$ to $\pi$. It's easy to see that $f$ is an injective immersion, but $f$ is not an embedding, since every neighborhood of $\mathbb{R}^2$ containing the origin also contains points of the form $f(-\pi + \epsilon)$ and $f(\pi - \epsilon)$ for all sufficiently small $\epsilon$.
As another one-dimensional example of this type, you could consider the closed topologist's sine curve with a loop, which is the graph of $g(x) = \sin(1/x)$ for $x \in (0, 1]$ together with the $y$-axis between $y = -1$ and $y = 1$ together with a "loop" smoothly connecting the point $(0, -1)$ to the point $(1, \sin(1))$. It's clear that there is some injective immersion $f : [0, \infty) \to \mathbb{R}^2$ whose image is this curve, and this immersion is not an embedding.
You can, of course, easily make either of these example into a surface by considering $h: (-\pi, \pi) \times (0,1) \to \mathbb{R}^3$ given by $h(t, s) = (f(t), s)$.
On the other hand, I don't really know any examples of injective immersions of surfaces which aren't embeddings that are "interesting" in a way that's fundamentally different from the examples above. The idea I have is that immersions are allowed to "approach themselves" or "limit onto themselves" in crazy ways that embeddings are not. In particular, if $f : X \to Y$ is an injective immersion, the topology on $f(X)$ as a subspace of $Y$ might be very different than the topology on $X$.
Consider any geodesic of unit speed $ \gamma :[0,\epsilon )
\rightarrow M$ If $\gamma(t)$ goes to $q\in M$, then by considering
$B_\delta (q)$ we can extend the geodesic $\gamma : [0,\epsilon
+\delta )\rightarrow M$
If we can do same thing countably many and if we have $\gamma : [0, \infty)
\rightarrow M$, then we have a completeness
And assume that $M$ is not complete So we have $\gamma : [0,\epsilon )\rightarrow M$ which
can not be extended.
By reparametrization we have $\alpha(t) =\gamma (\epsilon
(1-e^{-t})),\ t\in [0,\infty)$
If $
\alpha (n)$ are in some compact set $K$ so it has a limit in $K$. It is a contradiction since $\gamma$ can not
be extended. So any compact set does not contain all $\alpha (n)$ That is $\alpha$ is divergent curve
If $\alpha$ has infinite length then it is a contradiction since $ {\rm length}\ \alpha =
{\rm length}\ \gamma =\epsilon$
Best Answer
Every immersion $E^n\to M$ lifts to an immersion $E^n\to \tilde{M}$ by the basic covering theory (since $E^n$ is simply-connected). Each totally geodesic isometric immersion $E^n\to \tilde{M}$ is an embedding by Cartan-Hadamard theorem (you just need the target to be a complete simply connected manifold $X$ of nonpositive curvature). C-H theorem tells you is that the exponential map $exp_p: T_pX\to X$ is a diffeomorphism (for every base-point $p$). In particular, every nonconstant geodesic map ${\mathbb R}\to X$ is a (proper) embedding. Injectivity of $E^n\to X$ is now immediate.