Let $U$ denote the space we're working on and $\{U_i\}$ an open cover of $U$ (obviously $U$ may be an open set of an ambient space, but that plays no importance here). Let's assume that there is a sheaf $\mathscr F$ on $U$ and for each $i$ a section $s_i\in\mathscr F(U_i)$.

Gluing the $s_i$ means to find a section $s\in \mathscr F(U)$ such that for each $i$
$$ s|_{U_i}=s_i. $$

The obvious obstruction to this is that the $s_i$ and $s_j$ has to agree on the overlap. So, in order for such an $s$ to exist, we must have that
$$ (s_i)|_{U_{ij}}=(s_j)|_{U_{ij}}. \tag{$\star$}$$
In other words such that
the $0$-cochain $\{(U_i,s_i)\}$ is in $Z^0=H^0$ and this is clearly enough. One could argue, which might be your motivation, that this doesn't measure failure, but indeed it measures the success of gluing. I agree.

For $H^1$ I think it is still better to think of it as the measure of failure of *lifting*, but because we're talking about sheaves this means gluing in practice. In other words,
consider a surjective morphism of sheaves,
$$
\mathscr F \twoheadrightarrow \mathscr F'',
$$
and imagine wanting to lift sections of $\mathscr F''$ to $\mathscr F$. So, you start with a $t\in \mathscr F''(U)$ and from the sheaf properties you know that there exists $\mathfrak U=\{U_i\}$, an open cover of $U$, and for each $i$ a section $t_i\in\mathscr F(U_i)$ such that $ t|_{U_i}$ is the image of $t_i$ via the map $\mathscr F\to\mathscr F''$.

In order to lift $t$ to $\mathscr F(U)$ you need to glue the $t_i$. For that you consider the
the $1$-cochain $\sigma= \{(U_{ij},(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}})\}$.
Clearly $(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}}$ maps to $0$ in $\mathscr F''$, so it is naturally a $1$-cochain of sections of $\mathscr F'=\ker \big[\mathscr F\to\mathscr F''\big]$. An obvious computation shows that it is in $Z^1(\mathfrak U,\mathscr F')$.

Now observe that $\sigma\in B^1(\mathfrak U,\mathscr F')$ if and only if there exists a $0$-cochain
$\{(U_i,t_i')\}$ of $\mathscr F'$ such that
$$
(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}} = (t_i')|_{U_{ij}}-(t_j')|_{U_{ij}}
$$
which is the same as to say that the sections
$$
t_i-t_i'\in \mathscr F(U_i)
$$
satisfy $(\star)$ and hence they glue together to a section $t'\in\mathscr F(U)$. Notice, that
since $t_i'\in\mathscr F'(U_i)$, the image of $t_i-t_i'$ in $\mathscr F''(U_i)$ is the same as
the image of $t_i$, that is, $t|_{U_i}$.

In other words, the original $t\in \mathscr F''(U)$ can be lifted, or equivalently, the sections
$t_i-t_i'\in\mathscr F(U_i)$ can be glued if and only if the above associated $1$-cocycle in
$$
H^1(\mathfrak U, \mathscr F')=Z^1(\mathfrak U, \mathscr F')/B^1(\mathfrak U, \mathscr F')
$$
is $0$.

OK, so let's see about $H^2$. If you accept that $H^1$ is the obstruction to lifting sections, that is, lifting elements of $H^0$, then the same argument shows that $H^2$ measures the failure of lifting these obstructions.

Suppose you have a surjective map between two short exact sequences:
$$
\begin{matrix}
0 & & 0 \\
\downarrow & & \downarrow \\
\mathscr F' & \twoheadrightarrow & \mathscr G' \\
\downarrow & & \downarrow \\
\mathscr F & \twoheadrightarrow & \mathscr G \\
\downarrow & & \downarrow \\
\mathscr F'' & \twoheadrightarrow & \mathscr G'' \\
\downarrow & & \downarrow \\
0 & & 0 \\
\end{matrix}
$$
and let $\mathscr K'=\ker\big[ \mathscr F'\to \mathscr G'\big]$,
$\mathscr K=\ker\big[ \mathscr F\to \mathscr G\big]$, and
$\mathscr K''=\ker\big[ \mathscr F''\to \mathscr G''\big]$, so by the $9$ lemma there is a short exact sequence
$$
0 \to \mathscr K' \to \mathscr K\to \mathscr K'' \to 0
$$

Then as we found above, $H^1(\mathscr K'')$ measures the failure of lifting sections from $\mathscr G''$ (or equivalently gluing the local liftings) and $H^1(\mathscr K)$ measures the failure of lifting sections from $\mathscr G$ (or equivalently gluing the local liftings). The morphism induces a natural map
$$
H^1(\mathscr K)\to H^1(\mathscr K'')
$$
which is compatible with the above diagram in the sense that if $t\in \mathscr G(U)$ is a section, then the obstruction of lifting this to $\mathscr F(U)$ in $H^1(\mathscr K)$ maps to the obstruction
(in $H^1(\mathscr K'')$)
of lifting the image of $t$ in $\mathscr G''$ to $\mathscr F''$.

Now if you start with an element of $H^1(\mathscr K'')$, which could be an obstruction to lifting some section of $\mathscr G''$ to $\mathscr F''$, then the obstruction to lifting this to an element of $H^1(\mathscr K)$ lies in $H^2(\mathscr K')$. Of course, this is nothing else but saying in words what the long exact cohomology sequence means, but if you write down how these cohomology elements can be represented by a Cech cocycle and try to lift them by lifting locally and then gluing following the exact same steps as above, then you will get exactly this.

To get higher cohomology groups, you may iterate this process. Of course, it gets pretty hairy very soon.

Let me add that in my opinion the best way to understand higher cohomology is that it is the lower cohomology of syzygies. In other words, consider a sheaf $\mathscr F$ and embed it into an acyclic (e.g., flasque or injective or flabby or soft) sheaf. So you get a short exact sequence:
$$
0\to \mathscr F\to \mathscr A\to \mathscr G \to 0
$$
Since $\mathscr A$ is acyclic, we have that for $i>0$
$$
H^{i+1}(\mathscr F)\simeq H^i(\mathscr G),
$$
so if you understand what $H^1$ means, then $H^2$ of $\mathscr F$ is just $H^1$ of $\mathscr G$,
$H^3$ of $\mathscr F$ is just $H^2$ of $\mathscr G$ and so on.

This, of course, is not special to Cech cohomology, but you can use Cech cohomology to get a feeling for $H^1$ and then use this to get a feeling for $H^{>1}$.