How can we calculate the generalized gradient of $L^2\ni x\mapsto a\min(x(s),by(t))$? Let $(T,\mathcal T,\tau)$ be a measure space, $a,b\ge0$, $s,t\in T$ and $$f(x):=a\min(x(s),bx(t))\;\;\;\text{for }x\in L^2(\tau).$$

How can we calculate the generalized gradient $\partial_Cf(x)$ of $f$ at $x\in L^2(\tau)$?

We may note that $2\min(u,v)=u+v-|u-v|$ for all $u,v\in\mathbb R$. Now the only point on which $\mathbb R\ni u\mapsto|u|$ is not differentiable (in the classical sense) is $0$, but the generalized gradient at $0$ is easily seen to be $[-1,1]$. Moreover, $\mathbb R\setminus\{0\}\ni u\mapsto|u|$ is continuously differentiable and the generalized gradient at $x\in\mathbb R\setminus\{0\}$ is simply $\{x/|x|\}$. This knowledge should be helpful. However, I'm new to this topic and have no idea how to tackle the problem.
EDIT: As Iosif Pinelis pointed out, we need to assume that the singletons in $(T,\mathcal T,\tau)$ are measurable and admit positive measure to ensure that $f$ is locally Lipschitz continuous.
 A: $\newcommand{\de}{\delta}$
The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the evaluation functional $L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic. 

However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if $\tau(\{r\})>0$, then the evaluation functional $\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$ and $\tau(\{t\})>0$, we see that your nonlinear function(al)
\begin{equation*}
 f=\min(\de_s,b\de_t)
\end{equation*}
is Lipschitz. (I am assuming $a=1$, without loss of generality.) 
Moreover, then the generalized (upper) directional derivative of $f$ at $x\in L^2(\tau)$ in the direction $v\in L^2(\tau)$ is 
\begin{multline*}
 f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ 
 =\left\{\begin{aligned}
 v(s)&\text{ if }x(s)<bx(t), \\
 bv(t)&\text{ if }x(s)>bx(t), \\
  \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); 
 \end{aligned}\right. \tag{1}
\end{multline*}
see details on this at the end of this answer; here I am using the (somewhat strange to me) notation $f^0(x;v)$ from the linked paper by Clarke. 
So, the generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.
Consider now the case $x(s)=bx(t)$:  
Consider first the subcase $s\ne t$. We have 
$$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t
$$
iff $0\le A\le1$ and $B=(1-A)b$. So, if $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula 
$$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau)
$$
with $A\in[0,1]$. 
If $s=t$, $\tau(\{t\})>0$, and $x(t)=bx(t)$ (that is, either $b=1$ or $x(t)=0$), then 
$f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula 
$$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau)
$$
with $A\in[\min(1,b),\max(1,b)]$. 
Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that 
$f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, 
and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. 
Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$. 
To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that 
\begin{equation*}
 f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2}
\end{equation*}
On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, 
\begin{multline*}
 f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \\ 
 \le\max[hv(s),bhv(t)],
\end{multline*}
by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. 
So, 
\begin{equation*}
 f^0(x;v)\le\max[v(s),bv(t)]. \tag{3}
\end{equation*}
Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$. 
