Partial derivatives when the function's inputs are dependent on the same variable at 53:13 of this lecture by mit ocw, the prof. Moungi Bowendi writes,
$$ (\frac{\partial U}{\partial T})_{p} =  (\frac{\partial U}{\partial T})_{v} + (\frac{\partial U}{\partial T})_{T} (\frac{\partial V}{\partial T})_{p}$$
Before this he wrote,
$$ U(T,V(p,T) )$$
i.e: U is a function of both temperature and volume with volume being dependent on temperature and pressure. Now if this is so, how is it possible to take partial derivative of internal energy with respect to temperature at fixed volume because as soon as you change the temperature aren't you changing the volume?
 A: Initially you're considering $U$ as a function of $T,V$, i.e you have a function $U:\Bbb{R}^2\to \Bbb{R}$, $(T,V)\mapsto U(T,V)$. Physically you interpret this as telling you the internal energy of a system in terms of the temperature and volume.
Next, you have volume as a function of temperature and pressure. Formally, this means you have a function $\xi:\Bbb{R}^2\to \Bbb{R}$, where the physical interpretation is that for every $(p,T)\in \Bbb{R}^2 = \text{domain}(\xi)$, the value $\xi(p,T)\in \Bbb{R}$ gives the volume of the system when it is at a pressure $p$ and temperature $T$. And just so we're crystal clear on the notation, $\xi(a,b)$ is a real number which tells you the volume of the system when it is at a pressure of $a$ and at temperature $b$ (in whatever units you fancy).
Finally, you're constructing a new function which you're obtaining via composition, $\zeta:\Bbb{R}^2\to \Bbb{R}$ defined as
\begin{align}
\zeta(p,T):= U(T, \xi(p,T)).
\end{align}
Now, you have to apply the chain rule, which says that for all $(p,T)\in \Bbb{R}^2$,
\begin{align}
(\partial_2\zeta)_{(p,T)} &= (\partial_1U)_{(T,\xi(p,T))} + (\partial_2U)_{(T,\xi(p,T))} \cdot (\partial_2\xi)_{(p,T)}, \tag{i}
\end{align}
where I'm using the notation $(\partial_if)_{(\alpha,\beta)}$ to mean the partial derivative of the function $f$ with respect to it's $i^{th}$ entry, evaluated at the point $(\alpha,\beta)$.
Now, another way to write this same equation is to say that for all $(p,T) \in \Bbb{R}^2$,
\begin{align}
\dfrac{\partial \zeta}{\partial T}\bigg|_{(p,T)} &= \dfrac{\partial U}{\partial T}\bigg|_{(T, \xi(p,T))} + \dfrac{\partial U}{\partial V}\bigg|_{(T, \xi(p,T))} \cdot \dfrac{\partial \xi}{\partial T}\bigg|_{(p,T)} \tag{ii}
\end{align}

Throughout this answer I have intentionally used weird letters like $\xi, \zeta$ to emphasize that there are several different functions involved, and your confusion comes from the fact that you don't realize this (because the notation you use hides this fact, so it's very difficult to decipher unless you already know what you're doing).
The function $(p,T)\mapsto \xi(p,T)$ is physically what we interpret as "volume expressed as a function of pressure and temperature". You have denoted this as $V(p,T)$. I intentionally used the weird letter $\xi$ in order to distinguish between the function vs the letter $V$ you use in order to denote an input of the function $U$.
Similarly, the function $(p,T)\mapsto \zeta(p,T)$ is physically what we interpret as "internal energy expressed as a function of pressure and temperature". I intentionally used the weird letter $\zeta$ in order to distinguish between the $\zeta$ and $U$. Even though these have the same physical interpretation as giving the internal energy of the system, mathematically these are very different functions. We obtain $\zeta$ from $U$ via composition; we're composing $U$ with the function $(p,T)\mapsto (T,\xi(p,T))$ in order to get $\zeta$.
In the equation (note that your post has a typo)
\begin{align}
\left(\dfrac{\partial U}{\partial T}\right)_p &= \left(\dfrac{\partial U}{\partial T}\right)_V + \left(\dfrac{\partial U}{\partial V}\right)_T \cdot \left(\dfrac{\partial V}{\partial T}\right)_p, \tag{iii}
\end{align}
the notation $\left(\frac{\partial U}{\partial T}\right)_p$ means consider "$U$ as a function of $p,T$, and then differentiate with respect to $T$ while keeping $p$ fixed". But now you may be wondering that initially, $U$ was a function of $T,V$ so how is this possible? Well the answer is that if you interpret this statement literally like a robot (which is unfortunately how I interpreted it for most of my introductory thermodynamics course) then it's a completely nonsensical statement. How can $U$ be a function of $T,V$ initially and then suddenly become a function of $p,T$ later? The resolution to this "paradox" is to realize that we're talking about completely different mathematical functions $U$ vs $\zeta$.
One final remark is that people are lazy with regards to introducing new letters for new functions (especially in the context of chain rule), so that they'll use a letter like $V$ to mean both an independent variable and also to mean a function, or the same letter $U$ can mean two different functions. This double usage can be very confusing if you're unaware that this is being done... the only way around this is to practice writing statements in absolutely crystal clear unambiguous notation and then see how this relates to the more common statements.
For instance, (i) is the most unambiguous statement, and it is impossible to misinterpret (I honestly can't see anyway to misinterpret it). Next, (ii) is about as clear as (i), and really it's the best you can do with Leibniz's notation. Finally, you have (iii), which is easy to misinterpret for beginners because it repeatedly uses the same letter $U$ to mean two different functions, and the $U$ on the LHS has a completely different meaning from the $U$ on the RHS. You just have to practice until all three versions of the statement become just as obvious, so that you can translate back and forth between any of the notations.
