I want to prove that, if $f\colon \mathbb{R}\to \mathbb{R}$ is continuously differentiable except at $x_1,\dots, x_m$, where it is discontinuous, then $$f'=\frac{df}{dx}+\sum_{i=1}^m[f(x_i^+)-f(x_i^-)]\delta_{x_i}.$$
Of course the LHS is the weak derivative but I don't really know what is meant by $\frac{df}{dx}$ is this context since $f$ is not $C^1(\mathbb{R})$. How do I go about this? Of course I'm trying to proceed by the definition of weak derivative but don't really know how to proceed. My work \begin{align}\langle\frac{df}{dx}+\sum_{i=1}^m[f(x_i^+)-f(x_i^-)]\delta_{x_i},\varphi\rangle&=\langle\frac{df}{dx},\varphi\rangle+\sum_{i=1}^m[f(x_i^+)-f(x_i^-)]\langle\delta_{x_i},\varphi\rangle\\ &=\langle\frac{df}{dx},\varphi\rangle+\sum_{i=1}^m[f(x_i^+)-f(x_i^-)]\varphi(x_i)\end{align}
but I have no idea how to construct $-\langle f,\varphi'\rangle$ from this.
Without loss of generality, assume that $x_k<x_{k+1}$, i.e. the terms are in increasing order. Assume that $f$ is $C^1$ on the open intervals, $(-\infty,x_1)$, $(x_m,+\infty)$, and $(x_k,x_{k+1})$ for all $k=1,\dots,m-1$, and that $f$ is continuous on the closed intervals $(-\infty,x_1]$, $[x_m,+\infty)$, and $[x_k,x_{k+1}]$ for all $k=1,\dots,m-1$.
To explain the noation, on the left hand side the function $f$ has been identified with the distribution $T_f$ given by $$\langle T_f ,\varphi \rangle = \int_{\mathbb R} f(x) \varphi (x) $$ for all $\varphi \in C^\infty_0(\mathbb R)$. Then $f'$ means the distributional derivative of $T_f$. Now on the right hand side, $\frac{df}{dx}$ simply means the usual pointwise derivative of $f$ - there's no issue with this since $f$ is differentiable everywhere except on $\{x_1,\dots,x_m\}$ which has measure zero in $\mathbb R$.
Now to answer your question, recall that distributional derivative $T'$ of a distribution $T$ is, by definition, given by $\langle T', \varphi \rangle = - \langle T,\varphi'\rangle $. Hence, for all $\varphi \in C^\infty_0(\mathbb R)$,
\begin{align*} \langle f',\varphi \rangle = \langle T_f', \varphi \rangle &= - \langle T_f,\varphi'\rangle \\ &=- \int_{\mathbb R} f(x) \varphi'(x) \, dx \\ &=-\int_{-\infty}^{x_1} f(x) \varphi'(x) \, dx -\sum_{k=1}^{m-1} \int_{x_k}^{x_{k+1}} f(x) \varphi'(x) \, dx -\int_{x_m}^{+\infty} f(x) \varphi'(x) \, dx \end{align*} Next, we will integrating by parts on each interval. Note that $\varphi$ has compact support so the terms at $\pm \infty$ vanish and $f$ is possible discontinuous across each $x_k$, so we need to keep a track of which direction we approach from (notated by $f(x_k^+)$ and $f(x_k^-)$: \begin{align*} \langle f', \varphi \rangle &=\int_{-\infty}^{x_1} \varphi(x) \frac{df}{dx}(x) \, dx -f(x_1^-)\varphi(x_1) \\ &\qquad +\sum_{k=1}^{m-1} \bigg [ \int_{x_k}^{x_{k+1}} \varphi(x) \frac{df}{dx}(x) \, dx -f(x_{k+1}^-)\varphi(x_{k+1}) + f(x_k^+)\varphi(x_k)\bigg ] \\ &\qquad +\int_{x_m}^{+\infty} \varphi(x) \frac{df}{dx}(x) \, dx+ f(x_m^+)\varphi(x_m). \end{align*} Then expand the brackets and change dummy variable for the terms involving $f(x_{k+1}^-)\varphi(x_{k+1})$ to get\begin{align*} \langle f', \varphi \rangle &=\int_{\mathbb R}\varphi(x) \frac{df}{dx}(x) \, dx +\sum_{k=1}^m \big ( f(x_k^+)-f(x_k^-)\big )\varphi(x_k)\\ &= \bigg \langle \frac{df}{dx} + \sum_{k=1}^m \big ( f(x_k^+)-f(x_k^-)\big )\delta_{x_k}, \varphi \bigg \rangle \end{align*} using that $\langle \delta_a , \varphi \rangle = \varphi(a)$. Hence, $$f' =\frac{df}{dx} + \sum_{k=1}^m \big ( f(x_k^+)-f(x_k^-)\big )\delta_{x_k}.$$