Have I understood the multidimensional Itô formula correctly?

Question

I have seen the multidimensional Itô formula for continuous semi-martingales:

Fix $d\in \Bbb{N}$ and let $X=M+A$ be an $\Bbb{R}^d$ dimensional continuous semi-martingale. Let $f:\left[0,\infty\right)\times \Bbb{R}^d\to \Bbb{R}$ be a map which is once continuously differentiable in the first coordinate and twice continuously differentiable in the second coordinate. Then $$f(t,X_t)=f(0,X_0)+\int_o^t \partial_t f(s,X_s)~ds+\sum_{k=1}^d\int_0^t \partial_{x^k}f(s,X_s) ~dX_s^k+\frac{1}{2}\int_0^t\sum_{k,l=1}^d\partial^2_{x^kx^l}f(s,X_s)~d[M^k,M^l]_s$$

Now I have some problems in understanding this formula.

1. First of all is it correct that $\partial_t f(s,X_s)=\left.\frac{\partial f}{\partial t}\right|_{(t,X_t)=(s,X_s)}$.

I was initially confused since I haven't seen the meaning of differentiating $f(s,X_s)$ w.r.t $t$ since there is no $t$ in $f(s,X_s)$. But if I understood it correctly I first differentiate $f(t,X_t)$ w.r.t $t$ and then put $(s,X_s)$ into it?

2. My second question is, is it true that if $X=B$ is a $1$-dimensional Brownian motion, then I get $$f(t,B_t)=f(0,B_0)+\int_o^t \partial_t f(s,B_s)~ds+\int_0^t \partial_{x}f(s,B_s) ~dB_s+\frac{1}{2}\int_0^t\partial^2_{x^2}f(s,B_s)~ds$$

Thanks a lot for your help.