It is known that given a SES of real vector bundles $$0\rightarrow E\rightarrow F\rightarrow G\rightarrow 0$$, there is a splitting $F\simeq E\oplus G$, if the base space $B$ is paracompact. We can do this by choosing a metric on $F$.
If $B$ is not paracompact, what is an explicit example of such sequences that don't split?
The original question was unclear as to whether $B$ was an arbitrary topological space, a topological manifold or a smooth manifold, and whether $E$, $F$ and $G$ were smooth or continuous vector bundles. In the first half of this answer, $B$ is a Hausdorff, nonparacompact, topological space and the vector bundles are continuous. Further down, below the horizontal line, I will build a more complicated example using smooth manifolds.
The first part of this answer is extracted from
Schröer, Stefan, Pathologies in cohomology of non-paracompact Hausdorff spaces, Topology Appl. 160, No. 13, 1809-1815 (2013). ZBL1282.55010.
Essentially, I have combined the standard argument that $\mathrm{Ext}^1(C^0(\mathbb{R}), C^0(\mathbb{R}))$ is the same as $H^1(C^0(\mathbb{R}))$ with Shroer's construction of a nonzero Cech class in Proposition 3.1, using the simplification from Remark 3.6. (Here $C^0(\mathbb{R})$ is the sheaf of continuous real valued funtions.)
As a set, let $B = \{ 0 \} \sqcup {\Big (} \mathbb{N} \times (0,1] {\Big )}$, where $\mathbb{N} = \{ 1,2,3, \cdots \}$. A subset $\Omega$ of $B$ is open if and only if
The intersection $\Omega \cap {\Big(} \{n \} \times (0,1] {\Big)}$ is open for each $n \in \mathbb{N}$ AND
either $0 \not \in \Omega$ or else, for all but finitely many $n$, we have $\{ n \} \times (0,1) \subset \Omega$.
Let $U_n = \{ n \} \times (0,1]$ and let $V = \{ 0 \} \cup {\Big(} \mathbb{N} \times (0,1) {\Big)}$. Then $V \cup \bigcup_n U_n$ is an open cover of $B$ which has no locally finite subcover.
We have $V \cap U_n = \{ n \} \times (0,1)$; let $y_n$ be the function which projects $V \cap U_n$ onto this second coordinate. We build a rank two vector bundle $F$ on $B$ by gluing $V \times \mathbb{R}^2$ to $U_n \times \mathbb{R}^2$ by the map $\left( \begin{smallmatrix} 1 & (1-y_n)^{-1} \\ 0 & 1 \end{smallmatrix} \right)$.
Clearly, $F$ sits in a short exact sequence $0 \to \mathbb{R} \to F \to \mathbb{R} \to 0$. Suppose that this sequence had a splitting. Write this splitting on the coordinate chart $U_n \times (0,1)$ as $\left( \begin{smallmatrix} \sigma_n \\ 1 \end{smallmatrix} \right)$ and on the chart $V$ as $\left( \begin{smallmatrix} \tau \\ 1 \end{smallmatrix} \right)$. So $\sigma_n$ and $\tau$ are continuous functions $U_n \to \mathbb{R}$ and $V \to \mathbb{R}$ and $\sigma_n - \tau = (1-y_n)^{-1}$ on $U_n \cap V$.
Since $\tau$ is continuous, there is an open neighborhood $\Omega$ of $0$ where $|\tau| < 1$. For all but finitely many $n$, the set $\{ n \} \times (0,1)$ is contained in this $\Omega$. So, for such $n$, the function $\tau$ is bounded as we approach $(n,1)$. But $\sigma_n$ is continuous on $U_n$, so it is also bounded as we approach $(n,1)$. This contradicts that $\sigma_n - \tau = (1-y_n)^{-1}$, which is unbounded near $(n,1)$.
I thought about this more, and I can make a variant of this construction with smooth manifolds, taking $B$ to be the Prufer surface. Recall that $B$ has an open cover consisting of sets $U_a \cong \mathbb{R}^2$ for each $a \in \mathbb{R}$ and also $H \cong \mathbb{R} \times \mathbb{R}_{>0}$, with the point $(x,y) \in U_a$ glued to $(a+xy, y)$ in $H$. In particular, note that $y$ is a well defined global smooth function on $B$.
Let $f: \mathbb{R} \to \mathbb{R}$ be a function which is not a pointwise limit of continuous functions. We build a rank two vector bundle $F$ over $B$ by gluing $U_a \times \mathbb{R}^2$ to $H \times \mathbb{R}^2$ by the map $\left[ \begin{smallmatrix} 1 & f(a) y^{-1} \\ 0 & 1 \end{smallmatrix} \right]$.
Once again, we have a short exact sequence $0 \to \mathbb{R} \to F \to \mathbb{R} \to 0$. Suppose that it were split, with the splitting given by $\left[ \begin{smallmatrix} \sigma_a \\ 1 \end{smallmatrix} \right]$ on $U_a$ and $\left[ \begin{smallmatrix} \tau \\ 1 \end{smallmatrix} \right]$ on $H$. Then $$\sigma_a(x,y) - \tau(a+xy,y) = f(a) y^{-1}$$ for $(x,y) \in \mathbb{R} \times \mathbb{R}_{>0}$. Also, $\sigma_a$ extends continuously to $y=0$, so $\lim_{y \to 0^+} \sigma_a(0,y) y = 0$. We deduce that $$\lim_{y \to 0^{+}} \tau(a,y) y = f(a).$$ But this exhibits $f$ as a pointwise limit of continuous functions, contradicting our hypothesis.
