The problem is that you do not have uniform convergence. The compactness argument is usually based on the boundedness of the $C^1$-norm (often you do not even have this, see the bubbling described below). The $C^1$-bound allows you to extract from the Arzelà-Ascoli theorem for any compact subset of the domain a subsequence that converges in $C^0$ (on the chosen domain!) If you choose a larger compact subdomain, you have to choose a subsequence of the subsequence, and this way it is not possible to get a subsequence converging uniformly (you only obtain locally uniform convergence).
Also you will need to choose a metric on the strip, and certainly whether you really get uniform boundedness or not, will also depend on the chosen metric.
I can try to explain how bubbling works on holomorphic curves with compact domain (let's say a sphere, which simplifies things a lot since there is only one complex structure, and there is no need to worry about Deligne-Mumford compactification etc.). Also since the domain is compact, we do not need to worry about the choice of the metric.
I hope that this allows you to see what you have to think about to understand the holomorphic strip.
Suppose you have a family $u_n\colon S^2 \to M$. There are two cases:
If the family $u_n$ is uniformly bounded in $C^1$,then by Arzelá-Ascoli we have a subsequence that converges in $C^0$ to a map $u_\infty\colon S^2 \to M$ (same domain!), and by elliptic regularity, the subsequence will even converge in $C^\infty$ and $u_\infty$ is $J$-holomorphic.
[Remark: Think about this as Morera's theorem from classical complex analysis. $C^0$-convergence guarantees the convergence of the integral, since all integrals are $0$, so must be their limit, thus $u_\infty$ is holomorphic.]
- If the family $u_n$ is not uniformly bounded in $C^1$,then (assuming that $M$ is compact, so that the $u_n$ cannot simply "run off"), there will be a sequence $z_n\in S^2$ such that the differentials $\|Du_n(z_n)\| \to \infty$ as $n\to \infty$. (strictly speaking you have to choose a subsequence etc. but I want to keep the notation simple).
This of course spoils uniform convergence, but we still can obtain some information by a process called rescaling.
We can choose a complex chart around each $z_n \in S^2$. Let's say $\phi_n\colon D^2\subset \mathbb{C} \to S^2$ such that $\phi_n(0) = z_n$. Assume that $\phi(D^2)$ are all disks of radius $1$ in $S^2$.
Compose
$$\hat u_n := u_n\circ \phi_n \colon D^2 \to M$$
to obtain a family of holomorphic disks where $R_n := \|D\hat u_n(0)\| \to \infty$.
The maps $\hat u_n$ do not convergence uniformly either, but we can rescale the domains by multiplying them with $R_n$ and define rescaled holomorphic maps as follows
$$\hat v_n\colon R_n\cdot D^2 \to M, \, z \mapsto \hat u_n(\frac{z}{R_n})$$
Observe that
- The domain $R_n\cdot D^2$ of the maps $\hat v_n$ increases and eventually exhaust $\mathbb{C}$ since $R_n\to \infty$.
- The differential $D\hat v_n$ has norm $\|D\hat v_n\| = \frac{1}{R_n} \cdot \|D\hat u_n|\|$, and if we have chosen the $z_n$ suitably, the norm of $\|D\hat v_n\|$ will be bounded by $1$ on all of $R_n\cdot D^2$.
Consequence: The $\hat v_n$ do not convergence uniformly but only locally uniformly, that is, fix any radius $R\gg 1$, then from some sufficiently large $n$ on, there will be a subsequence such that we find a subsequence such that $\hat v_n$ converges uniformly (first in $C^0$ and then by elliptic regularity in $C^\infty$) on the disk of radius $R$.
Unfortunately (and contrary to your sketch above), we cannot say what happens at infinity, we can only guarantee existence of a holomorphic plane $u_\infty\colon \mathbb{C} \to M$ that will be the local uniform limit.
What do you do next? Next, you use the removal of singularity theorem. It tells you that any map $u_\infty\colon \mathbb{C} \to M$ with finite energy extends to a holomorphic map $\hat u_\infty\colon \hat {\mathbb{C}} = S^2 \to M$, and you have recovered your first bubble.
The rest of the compactness theorem is unfortunately much more technical: You have shown that there are small neighborhoods of the points $z_n$ in $S^2$ that produce a holomorphic sphere. Next you have to study $S^2$ with small disks around the $z_n$ removed and think if this also converges, possibly repeating the above argument to recover more bubbles, and once you have everything under control, because you know that every bubble has a minimal energy and the sum of all bubbles cannot have more energy than your initial curves, you still have show that all bubbles "connect" to give a "tree".
Conclusion: Usually you do not start out with uniform convergence, but you obtain from the $C^1$-boundedness (possibly after rescaling) a subsequence that converges uniformly on a fixed compact domain. As you increase the domain you have to choose a further subsequence, and your limit curve will not be the uniform limit of any sequence, but only the uniform limit of a sequence after restricting to a compact subset.
