Let $\mathcal{M}$ be a smooth Riemannian manifold (not necessarily complete) and consider $\gamma : [0,1) \to \mathcal{M}$ a differentiable curve with bounded derivative, i.e., there exists constant $L>0$ such that
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\|\gamma'(\lambda)\|_{\gamma(\lambda)}< L$, $\quad\forall\lambda\in[0,1)$,
where $\|\cdot\|_{\gamma(\lambda)}$ is the norm induced by the Riemannian metric on $T_{\gamma(\lambda)}\mathcal{M}$.
From this, we can say that
$d(\gamma(0),\gamma(\lambda)) \leq \int_{0}^\lambda \|\gamma'(\tau\|_{\gamma(\tau)}\mathrm{d}\tau < \lambda L <L$,
where $d(\cdot,\cdot)$ is the Riemannian distance function on $\mathcal{M}$.
Therefore, the image of the curve is inside a ball of radius $L$ centered at the first point of the curve:
$\gamma(\lambda)\in B(\gamma(0),L) = \left\{ x\in \mathcal{M} : d(\gamma(0),x) < L\right\},\quad \forall\lambda\in[0,1)$.
Further assume that the ball $B(\gamma(0),2L)$ is entirely contained in $\mathcal{M}$.
Question : Is it true that $\lim_{\lambda\to1} \gamma(\lambda)$ exists and belongs to the manifold?
In $\mathbb{R}^n$, we can simply write $\gamma(\lambda) = \gamma(0) + \int_{0}^\lambda \gamma'(\tau)\mathrm{d}\tau$ and the boundedness of the tangent vector norm would allow us to take that integral to the limit $\lambda = 1$. But this way of writing the curve breaks down on the manifold.
Another approach that works for complete manifolds is to take a sequence of $\left\{\lambda_k\right\}\subset [0,1)$ such that $\lambda_k\to 1$ and by the boundedness of the tangent vector we can show that the sequence $\left\{\gamma(\lambda_k)\right\}\subset \mathcal{M}$ is a Cauchy sequence and thus converges to something inside the manifold.
I would really like not to invoke completeness and somehow use the assumption that the ball of radius $2L$ is in the manifold, which excludes that the limit point is no longer on the manifold (think of a manifold that is an open subset of a vector space and that the curve goes up to the boundary of the manifold).
Any idea or comment is appreciated.
Look at your definition of $B(\gamma(0),L)$: it's the set of points in $\boldsymbol{\mathcal M}$ whose distance from $\gamma(0)$ is less than $L$. Similarly, $B(\gamma(0),2L)$ is the set of points in $\mathcal M$ whose distance from $\gamma(0)$ is less than $2L$. Thus $B(\gamma(0),2L)$ is contained in $\mathcal M$ by definition.
The answer to your question is no, as shown by the following counterexample: Let $\mathcal M$ be the unit disk in $\mathbb R^2$ with its Euclidean metric, and let $\gamma:[0,1)\to \mathcal M$ be the unit-speed geodesic $\gamma(t) = (t,0)$. Then $\|\gamma'(t)\|$ is bounded by $1$, and $B(\gamma(0),2)$ is contained in $\mathcal M$ by the observation in the preceding paragraph, but $\lim_{t\nearrow 1}\gamma(t)$ does not exist in $\mathcal M$.
I think that perhaps what you have in mind is to consider geodesic balls instead of metric balls. A geodesic ball of radius $L$ around $p\in \mathcal M$ is a set of the form $\exp_p(B(0,L))$, provided that $\exp_p$ is defined on the entire set $B(0,L) = \{v\in T_p\mathcal M: \|v\|<L\}$ and is a diffeomorphism from that set to its image. What's true is that if there is a geodesic ball in $\mathcal M$ around $\gamma(0)$ of radius $2L$, then $\lim_{t\nearrow 1}\gamma(t)$ exists in $M$. The proof is based on the fact that the restriction of $\exp_p$ to the closed ball of radius $L$ around $0$ is a homeomorphism from a compact set onto its image, which is therefore compact, and $\gamma([0,1))$ is contained in this compact set.