ki3i:
A less heuristic proof is the following. Define the function $Y(t,T,\mathcal{P})$ such that, for each partition $\mathcal{P}$ (of size $n$) of the interval $[t,T]$, we have
$$
Y(t,T,\mathcal{P}) := -\sum\limits_{i=1}^{n} f(t,s_{i})(s_{i + 1} - s_{i}) = -\sum\limits_{i=1}^{n} f(t,s_{i})\Delta s_i\,.
$$
Observe that,
$$
\begin{eqnarray*}
\sum\limits_{j=1}^{n}\frac{\partial}{\partial f_{t, s_{j}}} Y(t,T,\mathcal{P}) ~\mathrm df(t,s_{j}) = -\sum\limits_{i=1}^{n} 1\cdot\mathrm df(t,s_{i})\Delta s_i\,.\tag{1}\newline
\sum\limits_{j=1}^{n}\frac{1}{2}\frac{\partial^2}{\partial f^2_{t, s_{j}}} Y(t,T,\mathcal{P}) ~\mathrm d\langle f\rangle_{t,s_{j}} = -\sum\limits_{i=1}^{n} 0\cdot \mathrm d\langle f\rangle_{t,s_{t_i}}\Delta s_i = 0\,.\tag{2}\newline
\sum\limits_{j<r=1}^{n}\frac{\partial^2}{\partial f_{t, s_{j}}\partial f_{t, s_{r}}} Y(t,T,\mathcal{P}) ~\mathrm d\langle f, f\rangle_{t,s_{j},s_{r}} = -\sum\limits_{i<r=1}^{n} 0\cdot ~\mathrm d\langle f, f\rangle_{t,s_{i},s_{r}}\Delta s_i = 0\,.\tag{3}
\end{eqnarray*}
$$
Therefore, by Ito's Lemma, $(1)$, $(2)$ and $(3)$ imply that
$$
\mathrm dY(t,T,\mathcal{P}) = \frac{\partial}{\partial t}Y(t,T,\mathcal{P})~\mathrm dt - \sum\limits_{i=1}^{n} \mathrm df(t,s_{t_i})\Delta s_i\,.
$$
This means that, for each partition $\mathcal{P}^{'}$ (of size $m$) of the interval $[0,t]$, we have
$$
\begin{array}{rcl}
\displaystyle \sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T, \mathcal{P}) &=& \displaystyle \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t}Y(s_k, T, \mathcal{P})\right)\Delta s_k - \sum\limits_{i=1}^{n} \left(\sum\limits_{k=0}^{m} \Delta f_k(s_k,s_{i})\right)\Delta s_i\,. \\
&&\\
\mbox{So, }\displaystyle \,\,\lim\limits_{\|\mathcal{P}\|\rightarrow 0}\sum\limits_{k=0}^{m} \Delta Y_{k} &=&\displaystyle \lim\limits_{\|\mathcal{P}\|\rightarrow 0} \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T, \mathcal{P})\right)\Delta s_k - \lim\limits_{\|\mathcal{P}\|\rightarrow 0}\sum\limits_{i=1}^{n} \left(\sum\limits_{k=0}^{m} \Delta f_k(s_k,s_{i})\right)\Delta s_i \\
&=&\displaystyle \sum\limits_{k=0}^{m}\frac{\partial}{\partial t} \left(\lim\limits_{\|\mathcal{P}\|\rightarrow 0} Y(s_k, T, \mathcal{P})\right)\Delta s_k - \sum\limits_{k=0}^{m} \Big(\int\limits_{t}^{T}\Delta f_k(s_k,s)~\mathrm ds\Big)_k \\
&=&\displaystyle \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\sum\limits_{k=0}^{m} \Big(\Delta f_k(s_k,s)\Big)_k~\mathrm ds \\
&=&\displaystyle \sum\limits_{k=0}^{m} \left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds\,. \\
\therefore\,\, \sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T) &=&\displaystyle \sum\limits_{k=0}^{m} \left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds \\
&&\\
\mbox{Consequently, }\quad\quad&& \\
Y(t,T) -Y(0,T)&=&\displaystyle \lim\limits_{\|\mathcal{P^{'}}\|\rightarrow 0}\sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T) \\
&=&\displaystyle \lim\limits_{\|\mathcal{P^{'}}\|\rightarrow 0}\sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds \\
&=&\displaystyle \int\limits_{0}^{t} \left(\frac{\partial}{\partial t}Y(s,T)\right)\mathrm ds - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds\,.
\end{array}
$$
Or, if you prefer the SDE form,
$$
\mathrm dY(t,T) = \Big(\frac{\partial}{\partial t} Y(t,T)\Big)\mathrm dt - \int\limits_{t}^{T} \Big(\mathrm df(t, s)\Big) ~\mathrm ds\,.
$$
Admittedly, this is not as "air tight" as it can be. For instance, above, I assumed that as a differentiable function of $t$ and $s$ (not a process), $f(t,s)$ is sufficiently smooth to allow the interchange of the limiting operations "$\frac{\partial}{\partial t}$" and "$\|{\mathcal{P}}\|\rightarrow 0$", and that this is sufficient for the pertinent Ito-integrals to be well-defined and agree.